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;
#define rep(i,n) for(int i = 0; i < (int)n; i++)
using ll = long long;
const int MAX = 510000;
const ll mod = 1e9+7;

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

void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i-1] * (ll)i % mod;
        inv[i] = mod - inv[mod%i] * (mod/(ll)i) % mod;
        finv[i] = finv[i-1] * inv[i] % mod;
    }
}

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

int main(){
  ll n, k;
  cin >> n >> k;
  ll a[n];
  rep(i,n) cin >> a[i];
  sort(a,a+n);
  ll ans = 0;
  COMinit();
  for(int i = 1; i <= n; i++) {
    (ans += COM(i-1,k-1)*a[i-1]) %= mod;
    (ans -= COM(n-i,k-1)*a[i-1]) %= mod;
  }
  cout << (ans+mod)%mod << endl;
  return 0;
}