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 N=1e5+5;
const int MD=1e9+7;
int a[N],fac[N],inv[N];
int quick_pow(int x,int y) {
	int ans=1;
	while(y) {
		if(y&1) ans=1LL*ans*x%MD;
		y>>=1;
		x=1LL*x*x%MD;
	}
	return ans;
}
int C(int n,int m) {
	if(n<m) return 0;
	return 1LL*fac[n]*inv[m]%MD*inv[n-m]%MD;
}
int add(int &x,int y) {
	x+=y;
	if(x>=MD) x-=MD;
	if(x<0) x+=MD;
}
int main() {
	fac[0]=1;
	for(int i=1;i<N;i++) fac[i]=1LL*fac[i-1]*i%MD;
	inv[N-1]=quick_pow(fac[N-1],MD-2);
	for(int i=N-2;i>=0;i--) inv[i]=1LL*inv[i+1]*(i+1)%MD;
	int n,k;
	scanf("%d%d",&n,&k);
	for(int i=0;i<n;i++) {
		scanf("%d",&a[i]);
	}
	sort(a,a+n);
	int ans=0;
	for(int i=0;i<n;i++) {
		add(ans,1LL*a[i]*C(i,k-1)%MD);
		add(ans,1LL*(-1)*a[i]*C(n-i-1,k-1)%MD);
	}
	printf("%d\n",ans);
	return 0;
}