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>

const int mod=1e9+7;
int n,k;
long long a[100010];
long long jc[100010]={1},inv[100010]={1};

long long pow(long long a,int b)
{
	long long ans=1;
	while(b)
	{
		if(b&1) ans=ans*a%mod;
		a=a*a%mod;
		b>>=1;
	}
	return ans;
}

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

int main()
{
	// freopen("test.in","r",stdin);
	scanf("%d%d",&n,&k);
	if(k==1)
	{
		puts("0");
		return 0;
	}
	for(int i=1;i<=n;i++)
		scanf("%lld",a+i),jc[i]=jc[i-1]*i%mod,inv[i]=pow(jc[i],mod-2);
	std::sort(a+1,a+1+n);
	for(int i=1;i<=n;i++) a[i]+=a[i-1],a[i]=(a[i]%mod+mod)%mod;
	long long ans=0;
	for(int i=k;i<=n;i++)
	{
		ans+=((a[n]-a[i-1]-a[n-i+1])%mod+mod)%mod*c(i-2,k-2)%mod;
		ans%=mod;
	}
	printf("%lld\n",ans);
	return 0;
}
