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
/*
	In the name of ALLAH
	Author : Raashid Anwar
*/

#include <bits/stdc++.h>
using namespace std;

#define int int64_t
#define mod 1000000007

int F[100001];

int power(int a,int b) {
	int r=1;
	while(b) {
		if(b&1)
			r=(r*a)%mod;
		a=(a*a)%mod;
		b>>=1;
	}
	return r;
}

int C(int a,int b) {
	if(a-b<0||a<0||b<0)
		return 0;
	int ans=(F[a-b]*F[b])%mod;
	ans=(power(ans,mod-2));
	ans=(ans*F[a])%mod;
	return ans;
}


int32_t main() {
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	F[0]=1;
	for(int i=1;i<=100000;i++)
		F[i]=(F[i-1]*i)%mod;
	int n,k,ans=0;
	cin>>n>>k;
	int A[n+1];
	for(int i=1;i<=n;i++)
		cin>>A[i];
	sort(A+1,A+n+1);
	for(int i=1;i<=n;i++) {
		if(i>=k)
			ans=(ans+((A[i]*C(i-1,k-1))%mod))%mod;
		if(i<=n-k+1)
			ans=(ans+mod-((A[i]*C(n-i,k-1))%mod))%mod;
	}
	cout<<ans<<"\n";
}
