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;
using ll = long long;
ll MOD = 1000000007;

ll power(ll x,int y){
  if(y==0){
    return 1;
  }else if(y==1){
    return x%MOD;
  }else if(y%2==0){
    return power(x,y/2)*power(x,y/2)%MOD;
  }else{
    return power(x,y-1)*x%MOD;
  }
}

ll no(ll x){
  if(x<0){
    x = MOD+(x%MOD);
  }
  x = x%MOD;
  return x;
}

int main(){
  ll N,K;
  cin >> N >> K;
  vector<ll> C(N+1,0);
  for(ll n=(K-1);n<=N;n++){
    if(n==(K-1)){
      C[n]=1;
    }else{
      C[n] = C[n-1]*n%MOD*power(n-(K-1),MOD-2)%MOD;
    }
  }
  //for(int i=0;i<=N;i++){
  //  cout << C[i] << " ";
  //}
  ll sum=0;
  vector<ll> A(N,0);
  for(int i=0;i<N;i++){
    cin >> A[i];
  }
  sort(A.begin(),A.end());
  for(ll i=0;i<N;i++){
    sum += no((C[i]-C[N-1-i])*A[i]);
    sum = sum%MOD;
  }
  cout << sum << endl;
}