Gym 100952 D. Time to go back

http://codeforces.com/gym/100952/problem/D

D. Time to go back

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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
商店有n件商品，有m个朋友。其中又有k个十分亲密的朋友，十分亲密的朋友的礼物价值不得低于d，显然是一个排列组合，但数据量大，构造组合函数精度容易出问题，因此可以用杨辉三角
vis[i][j]=vis[i-1][j-1]+vis[i-1][j];

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#include <map>
#include <vector>
#include <algorithm>
using namespace std;
#define lowbit(x) (x&(-x))
#define max(x,y) (x>y?x:y)
#define min(x,y) (x<y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define PI 3.141592653589793238462
#define INF 0x3f3f3f3f3f
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long ll;
ll vis[][],val[],n,m,k,d,t;
bool cmp(int a,int b)
{
    return a>b;
}
int main()
{
    mem(vis);
    for(int i=;i<;i++)
    {
        vis[i][]=;
        for(int j=;j<=i;j++)
        {
            vis[i][j]=(((vis[i-][j-])%MOD)+((vis[i-][j])%MOD))%MOD;
        }
    }
    scanf("%lld",&t);
    while(t--)
    {
        ll ans=,pos=;
        scanf("%lld%lld%lld%lld",&n,&m,&k,&d);
        for(ll i=;i<n;i++)
            scanf("%lld",&val[i]);
        sort(val,val+n,cmp);
        for(ll i=;i<n;i++)
        {
            if(val[i]>=d) ans++;
            else break;
        }
        if(n-ans==) pos=vis[ans][m];
        else if(ans<k || n<m) pos=;
        else
        {
            for(ll i=k;i<=ans;i++)
            {
                if(m-i<) break;
                pos=(pos+(vis[ans][i]*vis[n-ans][m-i])%MOD)%MOD;
            }
        }
        printf("%lld\n",pos);
    }
}