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 N,K,A[1<<17],M=1e9+7,F[1<<17],I[1<<17],i,r;
long C(long n,long k){return F[n]*I[k]%M*I[n-k]%M;}
long P(long n,long r){return r?P(n*n%M,r/2)*(r%2?n:1)%M:1;}
int main(){
	cin>>N>>K;
	F[0]=1;
	for(i=1;i<=N;i++)F[i]=F[i-1]*i%M;
	I[N]=P(F[N],M-2);
	for(i=N;i--;)I[i]=I[i+1]*(i+1)%M;
	for(i=0;i<N;i++)cin>>A[i];
	sort(A,A+N);
	for(i=K-1;i<N;i++){
		(r+=A[i]*C(i,K-1))%=M;
		(r-=A[N-i-1]*C(i,K-1))%=M;
	}
	cout<<r%M<<endl;
}
