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;

#define rep(i, n) for (int i = 0; i < (int)(n); ++i)

const int mod = 1e9 + 7;

ll modpow(ll k, ll n) {
  ll res = 1;

  while (n > 0) {
    if (n & 1) res = res * k % mod;
    k = k * k % mod;
    n /= 2;
  }

  return res;
}

int main() {
  int n, k;
  cin >> n >> k;
  int a[n];
  rep(i, n) cin >> a[i];

  ll c[n];
  c[0] = 0;
  rep(i, n - 1) {
    if (i + 1 < k - 1)
      c[i + 1] = 0;
    else if (i + 1 == k - 1)
      c[i + 1] = 1;
    else
      c[i + 1] = (c[i] * (i + 1) % mod) * modpow(i - k + 2, mod - 2) % mod;
  }

  // rep(i, n) { cout << c[i] << endl; }

  sort(a, a + n);
  ll ans = 0;
  rep(i, n) {
    ans -= a[i] * c[n - i - 1] % mod;
    ans = (ans + mod) % mod;
    ans += a[i] * c[i] % mod;
    ans %= mod;
  }

  cout << ans << endl;

  return 0;
}
