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<bits/stdc++.h>
#include<unordered_map>
#define for_int(i,a,b) for(int i=(a);i<=(b);++i)
#define rep_int(i,a,b) for(int i=(a);i>=(b);--i)
#define for_ll(i,a,b) for(ll i=(a);i<=(b);++i)
#define rep_ll(i,a,b) for(ll i=(a);i>=(b);--i)
#define Ios ios::sync_with_stdio(false),cin.tie(0)
#define ull unsigned long long
#define ll long long
#define db double
#define HPMAX 1005
#define KMP_MAX 1005
#define prime_MAX 1000005
#define quick_pow_mod 1000000007
#define PII pair<int,int>
const int factor_N = 10000005;
const int tree_arr_N = 1000005;
using namespace std;
/*
	1.注意边界情况
	2.优先考虑时间复杂度
!!! 3.求最大值时，答案应至多初始化为INT_MIN；求最小值时，答案应至少初始化为INT_MAX
	4.遇事不决先暴力
	5.代码要写简洁
	6.计数题要开long long
*/
//快读
int read() {
	char c = getchar();
	int x = 0, f = 1;
	for (; !isdigit(c); c = getchar())
		if (c == '-') f = -1;
	for (; isdigit(c); c = getchar()) x = x * 10 + c - 48;
	return x * f;
}
int t, n, d;
int main(){
	t = read();
	while (t--) {
		n = read(); d = read();
		vector<int>a(n + 1, 0);
		for_int(i, 1, n) a[i] = read();
		sort(a.begin(), a.end());
		if (a[n] <= d || a[1] + a[2] <= d) printf("Yes\n");
		else printf("No\n");
	}

	
	return 0;
}
