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 long long long

const int MAX = 1e6;
const int MOD = 1e9+7;
long fac[MAX],finv[MAX],inv[MAX];

void combinit(){
    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;
    }
}


long com(int n, int k){
    if(n < k) return 0;
    if(n < 0 or k < 0) return 0;
    return fac[n] * (finv[k] * finv[n-k] % MOD) % MOD;
}

int main(){
  combinit();
  long N, K;
  cin >> N >> K;
  vector<long> A(N);
  for(int i=0; i<N; i++){
    cin >> A[i];
  }
  sort(A.begin(), A.end());
  long ans=0;
  for(int i=K-1; i<N; i++){
    ans = ans + com(i,K-1)*A[i]%MOD;
  }
  for(int i=0; i<N-K+1; i++){
    ans = ans - com(N-i-1, K-1)*A[i]%MOD;
  }
  cout << ans%MOD << endl;
  return 0;
}