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<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int N=100005,md=1000000007;
int n,K,i,a[N],jc[N],rjc[N],inv[N],ans;
int C(int n,int m)
{
	return 1ll*jc[n]*rjc[m]%md*rjc[n-m]%md;
}
int main()
{
	scanf("%d%d",&n,&K);
	inv[1]=1;
	for(i=2;i<=n;++i)
		inv[i]=1ll*inv[md%i]*(md-md/i)%md;
	jc[0]=rjc[0]=1;
	for(i=1;i<=n;++i)
		jc[i]=1ll*jc[i-1]*i%md,rjc[i]=1ll*rjc[i-1]*inv[i]%md;
	for(i=1;i<=n;++i)
		scanf("%d",a+i);
	sort(a+1,a+1+n);
	for(i=K;i<=n;++i)
		ans=(ans+1ll*(a[i]+md)%md*C(i-1,K-1))%md;
	for(i=1;i+K-1<=n;++i)
		ans=(ans+1ll*(-a[i]+md)%md*C(n-i,K-1)%md)%md;
	printf("%d",ans);
	return 0;
}
