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>
using namespace std;

typedef long long ll;
typedef vector<ll> vi;
typedef vector<vi> vvi;
typedef vector<string> vs;
typedef pair<ll,ll> pii;
typedef vector<pii> vpii;

#define pb(x) push_back(x)
#define ppb() pop_back()
#define mp(x,y) make_pair(x,y)
#define mod 1000000007
#define FOR0(i,n) for(ll i=0;i<n;i++)
#define FOR(i,a,b) for(ll i = a; i < b; i++)
#define FORD(i,a,b) for(ll i=(a);i>=(int)(b);i--)
#define sf(n) scanf("%lld", &n)
#define sfc(x) scanf("%c", &x)
#define pf(x) printf("%lld", x)
#define pff(x) printf("%lf", x)
#define pfc(x) printf("%c", x)
#define pfs(x) printf("%s", x)
#define sf2(a,b) scanf("%lld %lld", &a, &b)
#define sf3(a,b,c) scanf("%lld %lld %lld", &a, &b, &c)
#define pfn(n) printf("%lld\n", n)
#define pfs(n) printf("%lld ", n)
#define ff first
#define ss second   
#define mem(a,b) memset(a,b,sizeof(a))
#define sz size()
#define ll long long
#define all(c) (c).begin(),(c).end()
#define fast ios_base::sync_with_stdio(0); cin.tie(NULL); cout.tie(NULL);
const ll inf = 2e18;

bool prime[1000009];    
unordered_map<ll,ll> largest_factor;
void SieveOfEratosthenes(ll n)
{
    memset(prime,true, sizeof(prime)); 
    prime[1]=false;
    for (ll p=2; p*p<=n; p++)
    {
        if (prime[p] == true)
        {
            for (ll i=p*p; i<=n; i += p){
                prime[i] = false;
                largest_factor[i]=p;
            }
        }
    }
}


ll power(ll x,ll y)
{
    if (y == 0)
        return 1;
    else if (y%2 == 0)
        return power(x, y/2)*power(x, y/2);
    else
        return x*power(x, y/2)*power(x, y/2);
}


ll mod_power(ll a,ll b,ll m){
    ll res=1;
    a%=m;
    while(b){
        if(b&1){
            res=res*a%m;
        }
        a=a*a%m;
        b>>=1;
    }
    return res;
}



int main(){
    fast
    ll t;
    cin>>t;
    while(t--){
        ll n,d;
        cin>>n>>d;
        ll a[n];
        FOR0(i,n)cin>>a[i];
        sort(a,a+n);
        if(a[n-1]<=d || a[0]+a[1]<=d){
            cout<<"YES\n";
        }
        else cout<<"NO\n";
    }
} 



