A. How Much Does Daytona Cost?

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

We 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≤t≤10001≤�≤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≤n≤1001≤�≤100, 1≤k≤1001≤�≤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 a1�1, a2�2, a3�3, ……, an�� (1≤ai≤1001≤��≤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

Copy

7

5 4

1 4 3 4 1

4 1

2 3 4 4

5 6

43 5 60 4 2

2 5

1 5

4 1

5 3 3 1

1 3

3

5 3

3 4 1 5 5

output

Copy

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 44.

On the subsegment [2,5][2,5] the elements are 4, 3, 4, 14, 3, 4, 1.

• 44 appears 22 times;
• 11 appears 11 time;
• 33 appears 11 time.

This means that 44 is the most common element on the subsegment [2,5][2,5], so there exists a subsegment where 44 is the most common

解题说明：水题，只需要判断是否k出现过一次即可。

#include 
int main() 
{
	int t, n, k, a, f;
	scanf("%d", &t);
	while (t--)
	{
		scanf("%d %d", &n, &k);
		f = 0;
		while (n--)
		{
			scanf("%d", &a);
			if (a == k)
			{
				f = 1;
			}
		}
		if (f)
		{
			printf("YES\n");
		}
		else
		{
			printf("NO\n");
		}
	}
	return 0;
}