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;
long ncrmodp(long ans,long rkai,long nmrkai){

  long const mod = 1000000007;
  long tmp = (rkai*nmrkai)%mod;
  long i;

  for(i=1;i<=(mod-2);i*=2){
    if(i & (mod-2)){
      ans *= tmp;
      ans %= mod;
    }
    tmp = tmp*tmp;
    tmp%=mod;
  }
  return ans;
}

int main(){

  long const mod = 1000000007;
  int n,k;
  cin >> n >> k;
  vector<long>a(n);
  int i;
  for(i=0;i<n;i++){
    cin >> a[i];
  }

  vector<long>kai(n+1,1);

  for(i=1;i<=n;i++){
    kai[i] = kai[i-1]*i;
    kai[i]%=mod;
  }

  long ans = 0;
  sort(a.begin(),a.end());

  for(i=0;i<=n-k;i++){
    ans += ncrmodp(kai[n-1-i],kai[k-1],kai[n-k-i])*(a[n-1-i]-a[i]);
    ans %= mod;
  }
  while(ans<0)ans += mod*mod;
  ans %= mod;
  cout << ans << endl;

}
