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>
using namespace std;

const int MOD=(int)1e9+7;
const int MAX=510000;

long long fac[MAX],finv[MAX],inv[MAX];

//テーブルを作る前処理
void COMinit(){
  fac[0]=fac[1]=1;
  finv[0]=finv[1]=1;
  inv[1]=1;
  for(int i=2;i<MAX;i++){
    fac[i]=fac[i-1]*i%MOD;
    inv[i]=MOD-inv[MOD%i]*(MOD/i)%MOD;
    finv[i]=finv[i-1]*inv[i]%MOD;
  }
}

//二項係数の計算
long long COM(int n,int k){
  if(n<k) return 0;
  if(n<0 || k<0)  return 0;
  return fac[n]*(finv[k]*finv[n-k]%MOD)%MOD;
}

int main(){
	long long N,K;
	cin >> N >> K;
	vector<long long> A(N);
	for(int i=0;i<N;i++)	cin >> A.at(i);

	sort(A.begin(),A.end());
	long long ans=0;
	COMinit();
	for(int i=0;i<N;i++){
		ans-=(A.at(i)*COM(N-i-1,K-1))%MOD;
		//cout << i << "," << COM(N-i-1,K-1) << endl;
	}
	for(int i=N-1;i>=0;i--){
		ans+=(A.at(i)*COM(i,K-1))%MOD;
	}
	cout << ans%MOD << endl;

	return 0;
}
