You have an array a_1, a_2, ..., a_n. All a_i are positive integers.

In one step you can choose three distinct indices i, j, and k (i ≠ j; i ≠ k; j ≠ k) and assign the sum of a_j and a_k to a_i, i. e. make a_i = a_j + a_k.

Can you make all a_i lower or equal to d using the operation above any number of times (possibly, zero)?

Input

The first line contains a single integer t (1 ≤ t ≤ 2000) — the number of test cases.

The first line of each test case contains two integers n and d (3 ≤ n ≤ 100; 1 ≤ d ≤ 100) — the number of elements in the array a and the value d.

The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100) — the array a.

Output

For each test case, print YES, if it's possible to make all elements a_i less or equal than d using the operation above. Otherwise, print NO.

You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).

Example

Input


3
5 3
2 3 2 5 4
3 4
2 4 4
5 4
2 1 5 3 6


Output


NO
YES
YES

Note

In the first test case, we can prove that we can't make all a_i ≤ 3.

In the second test case, all a_i are already less or equal than d = 4.

In the third test case, we can, for example, choose i = 5, j = 1, k = 2 and make a_5 = a_1 + a_2 = 2 + 1 = 3. Array a will become [2, 1, 5, 3, 3].

After that we can make a_3 = a_5 + a_2 = 3 + 1 = 4. Array will become [2, 1, 4, 3, 3] and all elements are less or equal than d = 4.
#include <iostream>
#include <bits/stdc++.h>
using namespace std;
#include <vector>

// int fun()



int main() {
	// your code goes here
	int t;
	cin>>t;
	
	while(t--)
	{
		int n,d;
		cin>>n>>d;
		vector<int> arr(n,0);
		for(int i=0;i<n;i++)
		{
			cin>>arr[i];
			// cout<<arr[i];
		}
		
		sort(arr.begin(),arr.end());
		int i,j,k;
		i=n-1;
		j=0,k=j+1;
		while(arr[i]>d)
		{
			if((arr[j]+arr[k])<=d)
			{
				arr[i--] =arr[j]+arr[k];
				
			}
			else
				break;
		}
		if(arr[i]<=d)
			cout<<"YES"<<endl;
		else
			cout<<"NO"<<endl;
	}
	return 0;
}