Codeforces Round 900 (Div. 3) |
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Finished |

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A. How Much Does Daytona Cost?

time limit per test

1 secondmemory limit per test

256 megabytesinput

standard inputoutput

standard outputWe define an integer to be the most common on a subsegment, if its number of occurrences on that subsegment is larger than the number of occurrences of any other integer in that subsegment. A subsegment of an array is a consecutive segment of elements in the array $$$a$$$.

Given an array $$$a$$$ of size $$$n$$$, and an integer $$$k$$$, determine if there exists a non-empty subsegment of $$$a$$$ where $$$k$$$ is the most common element.

Input

Each test consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \le t \le 1000$$$) — the number of test cases. The description of test cases follows.

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 100$$$, $$$1 \le k \le 100$$$) — the number of elements in array and the element which must be the most common.

The second line of each test case contains $$$n$$$ integers $$$a_1$$$, $$$a_2$$$, $$$a_3$$$, $$$\dots$$$, $$$a_n$$$ ($$$1 \le a_i \le 100$$$) — elements of the array.

Output

For each test case output "YES" if there exists a subsegment in which $$$k$$$ is the most common element, and "NO" otherwise.

You can output the answer in any case (for example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as a positive answer).

Example

Input

75 41 4 3 4 14 12 3 4 45 643 5 60 4 22 51 54 15 3 3 11 335 33 4 1 5 5

Output

YES NO NO YES YES YES YES

Note

In the first test case we need to check if there is a subsegment where the most common element is $$$4$$$.

On the subsegment $$$[2,5]$$$ the elements are $$$4, \ 3, \ 4, \ 1$$$.

- $$$4$$$ appears $$$2$$$ times;
- $$$1$$$ appears $$$1$$$ time;
- $$$3$$$ appears $$$1$$$ time.

This means that $$$4$$$ is the most common element on the subsegment $$$[2, 5]$$$, so there exists a subsegment where $$$4$$$ is the most common element.

Codeforces (c) Copyright 2010-2023 Mike Mirzayanov

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