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+7,p=1e9+7;
int a[maxn],n,k;
long long fac[maxn],inv[maxn];
long long maxx,minn;

long long poww(int x,int y)
{
	long long ans=1,tmp=x;
	while(y) {
		if(y&1) ans=ans*tmp%p;
		tmp=tmp*tmp%p;
		y>>=1;
	}
	return ans;
}

long long c(int n,int m)
{
	return (fac[n]*inv[m]%p)*inv[n-m]%p;
}

int main()
{
	fac[0]=inv[0]=1;
	for(int i=1;i<maxn;++i) fac[i]=fac[i-1]*i%p;
	inv[maxn-1]=poww(fac[maxn-1],p-2);
	for(int i=maxn-2;i>=1;--i) inv[i]=inv[i+1]*(i+1)%p;
	scanf("%d %d",&n,&k);
	for(int i=1;i<=n;++i) scanf("%d",a+i);
	sort(a+1,a+n+1);
	for(int i=k;i<=n;++i) {
		maxx+=c(i-1,k-1)*a[i]%p;
		maxx%=p;
	}
	for(int i=1;i<=n-k+1;++i) {
		minn+=c(n-i,k-1)*a[i]%p;
		minn%=p;
	}
	printf("%lld",(maxx-minn+p)%p);
}