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;

long long MOD=1000000007;
long long expo(long long a,long long b){
	if(b==0)return 1;
	long long temp=expo(a,b/2);
	if(b%2==0)return (temp*temp)%MOD;
	return (((temp*temp)%MOD)*a)%MOD;
}
long long inv(long long a){
	return expo(a,MOD-2)%MOD;
}

long long fac[100005];

long long choose(long long n,long long k){
	return (((fac[n]*inv(fac[k]))%MOD)*inv(fac[n-k]))%MOD;
}

int main(){
	int n,k;
	scanf("%d%d",&n,&k);
	long long arr[n];
	for(int i=0;i<n;i++)scanf("%lld",&arr[i]);
	sort(arr,arr+n);
	fac[0]=1;
	for(long long i=1;i<100005;i++){
		fac[i]=fac[i-1]*i;fac[i]%=MOD;
	}
	long long ans=0;
	for(int i=0;i<n;i++){
		if(i<=n-k){
			ans-=(arr[i]*choose(n-i-1,k-1))%MOD;
		}
		if(i>=k-1){
			ans+=(arr[i]*choose(i,k-1))%MOD;
		}
		ans%=MOD;
		ans+=MOD;
		ans%=MOD;
	}
	printf("%lld",ans);
}
