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<cstring>
#include<algorithm>
#include<queue>
#include<map>
#include<set>
using namespace std;
#define LL long long 
#define LD long double
const int mod=1e9+7;
int n,k;
LL ans,a[100005],cheng[100005];
LL poww(LL x,LL nn)
{
	LL res=1;
	while(nn)
	{
		if(nn&1)res=res*x%mod;
		x=x*x%mod;
		nn>>=1;
	}
	return res;
}
LL cm(LL x,LL y)
{
	return cheng[x]*poww(cheng[y],mod-2)%mod*poww(cheng[x-y],mod-2)%mod;
}
int main()
{
	scanf("%d%d",&n,&k);
	for(int i=1;i<=n;i++)
		scanf("%lld",&a[i]);
	sort(a+1,a+n+1);
	cheng[0]=1;
	for(int i=1;i<=n;i++)
		cheng[i]=cheng[i-1]*i%mod;
	for(int i=1;i<=n-k+1;i++)
	{
		ans=(ans-a[i]*cm(n-i,k-1)%mod)%mod;	
	}
	for(int i=k;i<=n;i++)
	{
		ans=(ans+a[i]*cm(i-1,k-1)%mod)%mod;
	}
	cout<<(ans+mod)%mod<<endl;
	return 0;
}