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 ll long long

ll power(ll X,ll Y,ll P){
  vector<ll> digit2(1000),XX(1000);
  ll I,N,mod;
  I=0;
  
  while(Y!=0){
    digit2.at(I)=Y%2;
    Y=Y/2;
    I++;
  }
  N=I;
  
  XX.at(0)=X%P;
  for(I=1;I<N;I++){
    XX.at(I)=(XX.at(I-1)*XX.at(I-1))%P;
  }
  
  mod=1;
  for(I=0;I<N;I++){
    if(digit2.at(I)==1){
      mod=(mod*XX.at(I))%P;
    }
  }
  
  if(mod<0){
    mod=mod+P;
  }
  
  return mod;
}

int main(){
  ll n,k,i;
  cin >> n >> k;
  vector<ll> a(n);
  for(i=0;i<n;i++){
    cin >> a.at(i);
  }
  sort(a.begin(),a.end());
  
  ll p=1000000007;
  ll c=1,s=0;
  for(i=k-1;i<n;i++){
    s=(s+c*(a.at(i)-a.at(n-i-1)))%p;
    c=(((c*(i+1))%p)*power(i-k+2,p-2,p))%p;
  }
  
  if(s<0){
    s += p;
  }
  
  cout << s << endl;
}