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 rep(i,n) for(int i=0;i<(n);i++)
using ll=long long;
const ll MOD=1000000007;

int main() {
  ll N,K;
  cin>>N>>K;
  vector<ll> A(N);
  rep(i,N){
    cin>>A.at(i);
  }
  sort(A.begin(),A.end());
  vector<ll> F(N-K+1);
  rep(i,N-K+1){
    F.at(i)=A.at(K+i-1)-A.at(N-K-i);
  }//max-min,適切な係数をかける
  vector<ll> P(N-K+1);
  
  ll MAX=100000+10;
  vector<ll> fac(MAX), finv(MAX), inv(MAX);
  
  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;
  }
  
  rep(j,N-K+1){
    P.at(j)=fac.at(K+j-1)*finv.at(K-1)%MOD;
    P.at(j)*=finv.at(j);
    P.at(j)%=MOD;
  }
  ll ans=0;
  rep(j,N-K+1){
    ans+=F.at(j)*P.at(j)%MOD;
    ans%=MOD;
  }
  cout<<ans<<endl;
}



