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;
typedef long long ll;
#define rep(i,a) for(int i=0;i<(a);i++)
const ll MOD=1000000007;
//const ll MOD=998244353;

const int MAX_N=101010;
ll inv[MAX_N+1], fact[MAX_N+1], ifact[MAX_N+1];

ll comb(int n, int k){
  if(n<k) return 0;
  return fact[n]*ifact[k]%MOD*ifact[n-k]%MOD;
}

void prepare(){
  inv[1]=fact[0]=fact[1]=ifact[0]=ifact[1]=1LL;
  for(int i=2;i<=MAX_N;i++){
    inv[i]=inv[MOD%i]*(MOD-MOD/i)%MOD;
    fact[i]=fact[i-1]*i%MOD;
    ifact[i]=ifact[i-1]*inv[i]%MOD;
  }
}


int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  
  prepare();
  int N,K; cin>>N>>K;
  int A[N]; rep(i,N) cin>>A[i];
  sort(A,A+N);
  ll ans=0;
  for(int i=0;i<N;i++){
    if(i<=N-K){
      ans-=A[i]*comb(N-1-i,K-1)%MOD;
      ans+=MOD;
    }
    if(i>=K-1){
      ans+=A[i]*comb(i,K-1)%MOD;
      ans%=MOD;
    }
  }
  cout<<ans<<endl;
  return 0;
}