For a finite set of integers X, let f(X)=\max X - \min X.

Given are N integers A_1,...,A_N.

We will choose K of them and let S be the set of the integers chosen. If we distinguish elements with different indices even when their values are the same, there are {}_N C_K ways to make this choice. Find the sum of f(S) over all those ways.

Since the answer can be enormous, print it \bmod (10^9+7).

Constraints

* 1 \leq N \leq 10^5
* 1 \leq K \leq N
* |A_i| \leq 10^9

Input

Input is given from Standard Input in the following format:


N K
A_1 ... A_N


Output

Print the answer \bmod (10^9+7).

Examples

Input

4 2
1 1 3 4


Output

11


Input

6 3
10 10 10 -10 -10 -10


Output

360


Input

3 1
1 1 1


Output

0


Input

10 6
1000000000 1000000000 1000000000 1000000000 1000000000 0 0 0 0 0


Output

999998537
#include<bits/stdc++.h>

using namespace std;

const int maxn=1e5+5;

typedef long long ll;

const ll mod=1e9+7;

ll a[maxn];
ll inv[maxn];

ll qPow(ll a,int b){
	ll ans=1;
	while(b){
		if(b&1) ans=(ans*a)%mod;
		a=(a*a)%mod;
		b>>=1;
	}
	return ans;
}

int main(){
	int n,k;
	scanf("%d%d",&n,&k);
	for(int i=1;i<=n;i++){
		scanf("%lld",a+i);
	}
	sort(a+1,a+1+n);
	
	for(int i=1;i<=n;i++){
		inv[i]=qPow(i,mod-2);
	}
	
	ll ans=a[k];ll p=1;
	for(int i=k+1;i<=n;i++){
		p=(p*(i-1))%mod;
		p=(p*inv[i-k])%mod;
		ans=(ans+a[i]*p%mod)%mod;
	}
	ans=(ans+mod-a[n-k+1])%mod;
	p=1;
	for(int i=k+1;i<=n;i++){
		p=(p*(i-1))%mod;
		p=(p*inv[i-k])%mod;
		ans=(ans+mod-a[n-i+1]*p%mod)%mod;
	}
	printf("%lld",ans);
	return 0;
}