23:I[344796,["/_next/static/chunks/083cs_iw_ie0y.js","/_next/static/chunks/2sk2f47durg8o.js","/_next/static/chunks/00mfjk9z2ewie.js","/_next/static/chunks/2q9x6spyz4qxv.js","/_next/static/chunks/1wo5i8je1ygp5.js","/_next/static/chunks/05nh6118fe45j.js","/_next/static/chunks/2qcnia-og0qh0.js","/_next/static/chunks/3b7nv_cf1l4bq.js"],"default"] 27:I[27201,["/_next/static/chunks/083cs_iw_ie0y.js","/_next/static/chunks/2sk2f47durg8o.js","/_next/static/chunks/00mfjk9z2ewie.js","/_next/static/chunks/2q9x6spyz4qxv.js","/_next/static/chunks/1wo5i8je1ygp5.js","/_next/static/chunks/05nh6118fe45j.js","/_next/static/chunks/2qcnia-og0qh0.js"],"IconMark"] 24:Tb40,

You are given a list of strings. All strings in the list have the same length, and every string consists solely of uppercase English letters.

For each string, compute the difference pattern: the sequence of differences between consecutive letters, where each difference is the alphabetical position of the later letter minus that of the earlier letter (for example, 'B' - 'A' = 1 and 'C' - 'A' = 2, so a difference may be positive, negative, or zero).

In the list, all strings except exactly one share the same difference pattern. Identify and return the one string whose difference pattern is different from all the others.

Input Format

The first line contains a single integer n, the number of strings in the list.
Each of the next n lines contains one string.

Output Format

Print the single string whose difference pattern is unique (different from the pattern shared by every other string).

Constraints

3 <= n <= 26
2 <= length of each string <= 26
Every string consists of uppercase English letters (A-Z) only.
All strings in the list have the same length.
Exactly one string has a difference pattern that differs from the pattern shared by all the others; the answer is therefore always unique.

Example

Input

4
ACB
BDC
CED
DEF

Output

DEF

Explanation

Compute the difference pattern of each string (difference between each pair of adjacent letters):

ACB: (C - A, B - C) = (+2, -1)
BDC: (D - B, C - D) = (+2, -1)
CED: (E - C, D - E) = (+2, -1)
DEF: (E - D, F - E) = (+1, +1)

The first three strings share the difference pattern (+2, -1), while DEF has the different pattern (+1, +1). Since DEF is the odd one out, the answer is DEF.

Notes

A difference can be positive, negative, or zero (equal adjacent letters).
Two strings have the same difference pattern only if their difference sequences match position by position, including sign. AB (pattern (+1)) and BA (pattern (-1)) are different patterns.
Because every string has the same length, all difference patterns have the same length (length - 1).
Comparison is by the full pattern, not just the first difference.

25:T8d0,# Odd One Out Difference Pattern You are given a list of strings. All strings in the list have the same length, and every string consists solely of uppercase English letters. For each string, compute the **difference pattern**: the sequence of differences between consecutive letters, where each difference is the alphabetical position of the later letter minus that of the earlier letter (for example, `'B' - 'A' = 1` and `'C' - 'A' = 2`, so a difference may be positive, negative, or zero). In the list, **all strings except exactly one** share the same difference pattern. Identify and return the one string whose difference pattern is different from all the others. ## Input Format - The first line contains a single integer `n`, the number of strings in the list. - Each of the next `n` lines contains one string. ## Output Format Print the single string whose difference pattern is unique (different from the pattern shared by every other string). ## Constraints - `3 <= n <= 26` - `2 <= length of each string <= 26` - Every string consists of uppercase English letters (`A`-`Z`) only. - All strings in the list have the same length. - Exactly one string has a difference pattern that differs from the pattern shared by all the others; the answer is therefore always unique. ## Example **Input** ``` 4 ACB BDC CED DEF ``` **Output** ``` DEF ``` **Explanation** Compute the difference pattern of each string (difference between each pair of adjacent letters): - `ACB`: `(C - A, B - C) = (+2, -1)` - `BDC`: `(D - B, C - D) = (+2, -1)` - `CED`: `(E - C, D - E) = (+2, -1)` - `DEF`: `(E - D, F - E) = (+1, +1)` The first three strings share the difference pattern `(+2, -1)`, while `DEF` has the different pattern `(+1, +1)`. Since `DEF` is the odd one out, the answer is `DEF`. ## Notes - A difference can be positive, negative, or zero (equal adjacent letters). - Two strings have the same difference pattern only if their difference sequences match position by position, including sign. `AB` (pattern `(+1)`) and `BA` (pattern `(-1)`) are different patterns. - Because every string has the same length, all difference patterns have the same length (`length - 1`). - Comparison is by the full pattern, not just the first difference.26:T831,# Odd One Out Difference Pattern ## Overview You are given `n` equal-length uppercase strings. Every string except one shares the same "difference pattern" of gaps between adjacent letters, and you must return the single string whose pattern differs. ## Key Insight A string's identity for this problem is captured entirely by its difference pattern: the tuple of `s[i+1] - s[i]` values across adjacent letters. Two strings are considered equivalent here exactly when these tuples match position by position, including sign. So `AB` and `BA` are different even though they use the same letters. ## Approach Convert each string into its difference-pattern tuple. Count how many strings produce each distinct pattern. Because all strings but one share a common pattern, exactly one pattern occurs a single time. Scan the strings once more and return the first one whose pattern has a count of `1`. ## Why This Works The problem guarantees that all strings but exactly one share a pattern, so the multiset of patterns contains one pattern with multiplicity `n - 1` and one pattern with multiplicity `1`. The string mapping to the singleton pattern is uniquely the odd one out, so counting occurrences and returning the singleton is both correct and unambiguous. ## Complexity Let `n` be the number of strings and `L` their common length. Building every pattern costs `O(n * L)` time, and the counting map holds `O(n)` tuples of length `L - 1`, so total time is `O(n * L)` and space is `O(n * L)`. ## Common Pitfalls - Comparing only the first difference instead of the full pattern; two different patterns can share a first gap. - Using the absolute value of each difference, which wrongly merges patterns that are sign-flipped, such as `(+1)` and `(-1)`. - Treating the pattern as an unordered set or sorted list, which loses positional information. - Assuming a fixed string (for example the first one) is always part of the majority; the odd one out may appear at any position, including first or last. - Dropping the first or last adjacent difference through an off-by-one loop bound.1c:[["$","script",null,{"type":"application/ld+json","dangerouslySetInnerHTML":{"__html":"{\"@context\":\"https://schema.org\",\"@type\":\"LearningResource\",\"learningResourceType\":\"Coding Question\",\"name\":\"Odd One Out Difference Pattern\",\"description\":\"You are given a list of strings. 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