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<iostream>
#include<cstdio>
#include<algorithm>
#define p 1000000007
using namespace std;
int n,K,a[101010];
long long fac[101010],ny[101010],ans;
long long C(int x,int y){ return x<y?0:fac[x]*ny[y]%p*ny[x-y]%p;}
int main()
{
	scanf("%d%d",&n,&K);
	for(int i=1;i<=n;i++)
	scanf("%d",&a[i]);
	sort(a+1,a+n+1);
	fac[0]=ny[0]=ny[1]=1;
	for(int i=1;i<=n;i++) fac[i]=fac[i-1]*i%p;
	for(int i=2;i<=n;i++) ny[i]=ny[p%i]*(p-p/i)%p;
	for(int i=2;i<=n;i++) ny[i]=ny[i-1]*ny[i]%p;
	for(int i=1;i<=n;i++)
	(ans+=1LL*(C(i-1,K-1)-C(n-i,K-1)+p)%p*a[i]%p+p)%=p;
	printf("%lld\n",ans);
	return 0;
}