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>
#define fi first
#define se second
#define ins insert
#define pb push_back
#define eb emplace_back
#define sz(x) ((int)x.size())
#define all(x) x.begin(), x.end()
using namespace std;
typedef long long ll;
typedef pair<int, int> pii; //
void out(int a) {
	if (a) cout << "YES\n";
	else cout << "NO\n";
}
void solve() {
	int n, d;
	cin >> n >> d;
	vector<int> v(n);
	int is = 0;
	for (auto& i : v) {
		cin >> i;
		if (i > d) is = 1;
	}
	if (is == 0) {
		out(1);
		return;
	}
	if (n <= 2) { 
		out(0);
		return;
	}
	sort(all(v));
	out(v[0] + v[1] <= d);	
}
int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	int t;
	cin >> t;
	while(t--) {
		solve();
	}
	// cout << endl;
	// string sss;
	// while(cin>>sss);
	return 0;
}
