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;
const int mod=1e9+7;
const int nmax=1000010;
ll fac[nmax],finv[nmax],inv[nmax];
 
void COMinit(){
  fac[0]=fac[1]=1;
  finv[0]=finv[1]=1;
  inv[1]=1;
  for(int i=2;i<nmax;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;
  }
}
ll com(int n,int k){
  if(n<k||n<0||k<0){
    return 0;
  }
  return fac[n]*(finv[k]*finv[n-k]%mod)%mod;
}
 
int main(){
  COMinit();
  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 mx=0,mi;
  for(int i=0;i<n;i++){
    mx+=(com(i,k-1)*a[i])%mod;
    mi+=(com(n-1-i,k-1)*a[i])%mod;
    mx%=mod;
    mi%=mod;
  }
  cout << (mx+mod-mi)%mod << endl;
}