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 = 1e9+7;
ll fact[200005],factr[200005];
ll mod_pow(ll a, ll b){
  ll res = 1;
  while(b){
    if(b&1)res = res * a % mod;
    b /= 2;
    a = a * a % mod;
  }
  return res;
}
ll combi(int n, int k){
  if(n < 0 or k < 0 or n < k)return 0;
  return fact[n] * factr[n - k] % mod * factr[k] % mod;
}
int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  fact[0] = fact[1] = 1;
  for(int i = 0; i < 200000; i++){
    fact[i + 1] = fact[i] * (i + 1) % mod;
    factr[i] = mod_pow(fact[i], mod - 2);
  }
  int 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;
  for(int i = 0; i < n; i++){
    ans = (ans + a[i] * combi(i, k - 1) % mod) % mod;
    ans = (ans - a[i] * combi(n - i - 1, k - 1) % mod + mod) % mod;
  }
  cout << ans << endl;
}
