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<vector>
#include<algorithm>
using namespace std ;

int main(){
  using ll = long long int ;
  ll n,k ;
  cin >> n >> k ;
  vector<ll> kake(n+1) ;
  vector<ll> invs(n+1) ;
  vector<ll> invn(n+1) ;
  ll mod = 1000000007 ;
  vector<ll> a(n) ;
  for(int i=0;i<n;i++) cin >> a.at(i) ;
  kake.at(0) = kake.at(1) = 1 ;
  for(int i=2;i<=n;i++) kake.at(i) = kake.at(i-1)*i%mod ;
  invn.at(1) = 1 ;
  for(int i=2;i<=n;i++) invn.at(i) = mod-invn.at(mod%i)*(mod/i)%mod ;
  invs.at(0) = invs.at(1) = 1 ;
  for(int i=2;i<=n;i++) invs.at(i) = invs.at(i-1)*invn.at(i)%mod ;
  sort(a.begin(),a.end()) ;
  ll ans = 0 ;
  for(int i=0;i<n-k+1;i++){
    ll sub = a.at(n-1-i)-a.at(i) ;
    ll ncr = kake.at(n-1-i)*(invs.at(k-1)*invs.at(n-i-k)%mod)%mod ;
    ans += (sub*ncr%mod+mod)%mod ;
    ans %= mod ;
  }
  cout << ans << endl ;
}