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 int MAX = 5e5+3;
const ll mod = 1000000007;

ll fact[MAX], inv[MAX], finv[MAX];

ll nCk(ll n, ll k){
  if(n < k) return 0;
  if(n < 0 || k < 0) return 0;
  return fact[n] * finv[n-k] % mod * finv[k] % mod;
}

void pre_cal(){
  fact[0] = inv[1] = fact[1] = 1;
  finv[0] = finv[1] = 1;
  for(int i = 2; i <= 5e5; i++){
    fact[i] = fact[i-1] * i % mod;
    inv[i] = mod - (mod/i) * inv[mod%i] % mod;
    finv[i] = finv[i-1] * inv[i] % mod;
  }
}

int main(){
  pre_cal();
  ll n, k; cin >> n >> k;
  vector<ll> a(n);
  for(ll i = 0; i < n; i++) cin >> a[i];
  sort(a.begin(), a.end());
  ll mx = 0, mn = 0;
  for(ll i = k; i <= n; i++){
    mx += mod + nCk(i-1, k-1) * a[i-1] % mod;
    mx = (mx + mod) % mod;
  }
  for(ll i = 0; i <= n-k; i++){
    mn += nCk(n-i-1, k-1) * a[i] % mod;
    mn = (mn + mod) % mod;
  }
  cout << (mod + (mx - mn)%mod)%mod << endl;
}