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;
using ll = long long;
const ll mod = 1000000007;

int power(ll a, ll b){
 ll res=1;
 while(b>0){
   if(b & 1) res=res*a%mod;
    a=a*a%mod;
    b>>=1;
  }
return res;
}

int main(){
  ll n,k;
  cin >> n >> k;
  vector<ll>a(n);
  for(int i=0; i<n; i++) cin >>a[i];
  sort(a.begin(),a.end());
  ll ans=0,K=1;
  vector<ll>all(n+1,1);
  for(int i=1; i<=k-1; i++){
    K*=i;
    K%=mod;
  }
  all[k-1]=K;
  for(int i=k; i<=n; i++){
    all[i]=all[i-1];
    all[i]*=i;
    all[i]%=mod;
    all[i]*=power(i-k+1,mod-2);
    all[i]%=mod;
  }
  
  K=power(K,mod-2);
  
  for(int i=k-1; i<n; i++){
    ll x = a[i];
    x *=all[i];
    x%=mod;
    x *= K;
    x%=mod;
    ans+=x;
    ans%=mod;
  }
  
  for(int i=0; i<n-k+1; i++){
    ll y = a[i];
    y*=all[n-i-1];
    y%=mod;
    y*=K;
    y%=mod;
    ans-=y;
    ans%=mod;
  }
  
  if(ans<0)ans+=mod;
  
  
  cout << ans << endl;
  
  
  
}
  
