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>
#define N 1100000
using namespace std;
int a[N],i,j,n,m,k,l,o;
long long ans,now;
const int mo=1e9+7;
int ny(int x,int m)
{
	int ans=1,p=x;
	while (m)
	{
		if (m&1) ans=(1ll*ans*p)%mo;
		m>>=1;
		p=(1ll*p*p)%mo;
	}
	return ans;
}
int main()
{
	scanf("%d%d",&n,&k);
	for (i=1;i<=n;i++) scanf("%d",&a[i]);
	sort(a+1,a+n+1);
	now=1;
    for (i=k;i<=n;i++)
    {
    	ans=(ans+1ll*now*a[i])%mo;
    	now*=i;
    	now%=mo;
    	now*=ny(i-k+1,mo-2);
    	now%=mo;
	}
	now=1;j=k;
	for (i=n-k+1;i>=1;i--)
	{
		ans=(ans-1ll*a[i]*now)%mo;
		now*=j;
		now%=mo;
		now*=ny(j-k+1,mo-2);j++;
		now%=mo;
	}
	printf("%d",ans);
}