Gym

D. Time to go back

 

You have been out of Syria for a long time, and you recently decided to come back. You remember that you have M friends there and since you are a generous man/woman you want to buy a gift for each of them, so you went to a gift store that have N gifts, each of them has a price.

You have a lot of money so you don't have a problem with the sum of gifts' prices that you'll buy, but you have K close friends among your M friends you want their gifts to be expensive so the price of each of them is at least D.

Now you are wondering, in how many different ways can you choose the gifts?

Input

The input will start with a single integer T, the number of test cases. Each test case consists of two lines.

the first line will have four integers N, M, K, D (0  ≤  N, M  ≤  200, 0  ≤  K  ≤  50, 0  ≤  D  ≤  500).

The second line will have N positive integer number, the price of each gift.

The gift price is  ≤  500.

Output

Print one line for each test case, the number of different ways to choose the gifts (there will be always one way at least to choose the gifts).

As the number of ways can be too large, print it modulo 1000000007.

Examples

input

2
5 3 2 100
150 30 100 70 10
10 5 3 50
100 50 150 10 25 40 55 300 5 10

output

3
126

比赛时组合数求错了，网上看到一种更好的组合数求法，记一下。

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#define ll long long
using namespace std;
int Mod = 1000000007;
ll n,m,k,d;
ll a[220],C[220][220];
void getC(){
    memset(C,0,sizeof(C));
    for(int i = 0; i < 220; i ++){
        C[i][0] = 1;
        for(int j = 1; j <= i; j ++){
            C[i][j] = (C[i-1][j-1]+C[i-1][j])%Mod;
        }
    }
}
int main(){
    int t;
    getC();
    cin>>t;
    while(t--){
        cin>>n>>m>>k>>d;
        memset(a,0,sizeof(a));
        for(ll i = 1; i <= n; i ++)cin>>a[i];
        int cnt = count_if(a+1,a+n+1,bind2nd(greater_equal<int>(),d));
        ll ans = 0;
        for(ll i = m>=cnt?cnt:m; i >= k; i --){
            ans = (ans+(C[cnt][i]%Mod)*(C[n-cnt][m-i]%Mod))%Mod;
        }
        cout << ans%Mod << endl;
    }
    return 0;
}