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
const ll MOD=1e9+7;
vector<long long> fact, fact_inv, inv;
void init_nCk(ll SIZE){
  fact.resize(SIZE+5);
  fact_inv.resize(SIZE+5);
  inv.resize(SIZE+5);
  fact[0]=fact[1]=1;
  fact_inv[0]=fact_inv[1]=1;
  inv[1]=1;
  for(ll i=2;i<SIZE+5;++i){
    fact[i]=fact[i-1]*i%MOD;
    inv[i]=MOD-inv[MOD%i]*(MOD/i)%MOD;
    fact_inv[i]=fact_inv[i-1]*inv[i]%MOD;
  }
}
ll nCk(ll n,ll k){
  return fact[n]*(fact_inv[k]*fact_inv[n-k]%MOD)%MOD;
}
int main(){
  ll N,K,ans=0;
  cin>>N>>K;
  vector<ll> vec(N);
  for(ll i=0;i<N;i++){
    cin>>vec.at(i);
  }
  sort(vec.begin(),vec.end());
  if(K==1){cout<<0<<endl;}
  else{
    init_nCk(100000);
    for(ll i=K-1;i<N;i++){
      ans+=vec.at(i)*nCk(i,K-1)%MOD;
      ans=(ans+MOD)%MOD;
    }
    for(ll i=N-K;i>=0;i--){
      ans-=vec.at(i)*nCk(N-i-1,K-1)%MOD;
      ans=(ans+MOD)%MOD;
    }
    cout<<ans<<endl;
  }
}