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;

#define MOD 1000000007
#define MAX 2000000

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

void pre(){
  fac[0]=fac[1]=1;
  inv[1]=1;
  finv[0]=finv[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;
  }
}

long int com(int n,int k){
  return ((fac[n]*finv[k])%MOD)*finv[n-k]%MOD;
}

int main(){
  pre();
  int N,K;
  cin>>N>>K;
  vector<int> A(N);
  for(int i=0;i<N;i++){
    cin>>A.at(i);
  }
  
  sort(A.begin(),A.end());
  long int sum=0;
  for(int i=0;i<N-K+1;i++){
    sum+=A.at(i+K-1)*com(i+K-1,K-1)%MOD;
    sum-=(A.at(N-K-i)*com(i+K-1,K-1))%MOD;
    if(sum<0){
      sum+=MOD;
    }
    sum%=MOD;
  }
  cout<<sum<<endl;
}
  
  
  
  