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 <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;


int main(void){
  int n,k,i;
  ll a[100005];
  
  cin >> n >> k;
  for (i=0;i<n;i++) cin >> a[i];
  
  sort(a,a+n);
  
  ll fac[100005],inv[100005],finv[100005],ans=0;
  const ll mod=1e9+7;
  
  fac[0]=fac[1]=1;
  finv[0]=finv[1]=1;
  inv[1]=1;
  for (i=2;i<=n;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;
  }
  
  for (i=0;i<n;i++){
    if (i>=k-1) ans=(ans+fac[i]*finv[k-1]%mod*finv[i-k+1]%mod*a[i]%mod)%mod;
    if (i+k-1<n) ans=(ans+mod-fac[n-i-1]*finv[k-1]%mod*finv[n-i-k]%mod*a[i]%mod)%mod;
  }
  
  cout << ans << endl;
  
  return 0;
}