CodeChef

Read problems statements in Mandarin Chinese and Russian.

Little Elephant from Zoo of Lviv likes to watch movies.

There are N different movies (numbered from 0 to N − 1) he wants to watch in some order. Of course, he will watch each movie exactly once. The priority of i^th movie is P_i.

A watching of a movie is called exciting if and only if one of the following two conditions holds:

• This is the first watching.
• The priority of this movie is strictly greater than the maximal priority of the movies watched so far.

Let us call the number of exciting watchings the excitingness of the order.

Help him to find the number of different watching orders whose excitingness does not exceed K. Since the answer can be large, print it modulo 1000000007 (10⁹+7).

Input

The first line of input contains an integer T, denoting the number of test cases. Then T test cases follow.

The first line of each test case contains two space-separated integers N and K. The next line contains N space-separated integers P₁, P₂, ..., P_N.

Output

For each test case, print the number of different watching orders having at most Kexcitingness modulo 1000000007 (10⁹+7).

Constraints

• 1 ≤ T ≤ 10
• 1 ≤ K ≤ N ≤ 200
• 1 ≤ P_i ≤ 200

Example

Input:
2
3 1
3 1 2
4 3
1 2 2 3

Output:
2
24

Explanation

In the first case, there are two boring watching orders whose excitingness not greater than K=1: [3, 1, 2], [3, 2, 1]. Both watching orders have one excitingwatching: the first watching.

In the second case, every watching order has at most 3 excitingness.

    一般的排列问题都可以转化成把元素以一定顺序插入进序列来做。这个题如果按照升序来插入的话，后插入的元素会挡住前面插入的元素，没法算；

而如果按照降序插入的话，影响是很好计算的：插入的元素不会挡住前面的，而且只有放在最前面会对 激动值 +1。

    有很多重复的元素的话，我们就可重集组合一下，不过对激动值的 影响最大还是1，因为相同元素也会挡住。

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn=405;
const int ha=1000000007;
int jc[maxn],ni[maxn],T,n,k;
int dp[maxn],num[maxn],N,f[maxn];

inline int add(int x,int y){
	x+=y;
	return x>=ha?x-ha:x;
}

inline int ksm(int x,int y){
	int an=1;
	for(;y;y>>=1,x=x*(ll)x%ha) if(y&1) an=an*(ll)x%ha;
	return an;
}

inline void getC(){
	jc[0]=1;
	for(int i=1;i<=400;i++) jc[i]=jc[i-1]*(ll)i%ha;
	ni[400]=ksm(jc[400],ha-2);
	for(int i=400;i;i--) ni[i-1]=ni[i]*(ll)i%ha;
}

inline int C(int x,int y){ return x<y?0:jc[x]*(ll)ni[y]%ha*(ll)ni[x-y]%ha;}

inline void init(){
    memset(dp,0,sizeof(dp));
    memset(num,0,sizeof(num));
}

inline void solve(){
	dp[0]=1;
	for(int i=200,A=0,P=0,tot,lef;i;i--) if(num[i]){
		lef=C(A+num[i]-1,num[i])*(ll)jc[num[i]]%ha,tot=add(C(A+num[i],num[i])*(ll)jc[num[i]]%ha,ha-lef);
		
		memset(f,0,sizeof(f));
		for(int j=min(P,k);j>=0;j--){
			f[j+1]=add(f[j+1],dp[j]*(ll)tot%ha);
			f[j]=add(f[j],dp[j]*(ll)lef%ha);
		}
		memcpy(dp,f,sizeof(f));
		
		P++,A+=num[i];
	}
	
	int ans=0;
	for(int i=0;i<=k;i++) ans=add(ans,dp[i]);
	printf("%d
",ans);
}

int main(){
	getC();
	scanf("%d",&T);
	while(T--){
		init(),scanf("%d%d",&n,&k);
		for(int i=1;i<=n;i++) scanf("%d",&N),num[N]++;
		solve();
	}
	return 0;
}