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>
#define ll long long
#define MOD 1000000007

using namespace std;

ll modinv(ll a, ll m = MOD) {
  ll b = m, u = 1, v = 0;
  while(b) {
    ll t = a / b;
    a -= t * b, swap(a, b);
    u -= t * v, swap(u, v);
  }
  u %= m;
  if(u < 0) u += m;
  return u;
}

int main() {
  int n, k;
  ll ma = 0, mi = 0, f[100001];
  f[0] = f[1] = 1;
  for(int i = 2; i < 100001; i++) f[i] = f[i - 1] * i % MOD;
  scanf("%d %d", &n, &k);
  vector<int> a(n);
  for(int i = 0; i < n; i++) scanf("%d", &a[i]);
  sort(a.begin(), a.end());
  for(int i = k - 1; i < n; i++) {
    ma = (ma + f[i] * modinv(f[i - k + 1]) % MOD * modinv(f[k - 1]) % MOD *
                   a[i] % MOD) %
         MOD;
    mi = (mi + f[i] * modinv(f[i - k + 1]) % MOD * modinv(f[k - 1]) % MOD *
                   a[n - 1 - i] % MOD) %
         MOD;
  }
  ll o = ma - mi;
  while(o < 0) o += MOD;
  printf("%lld\n", o % MOD);
}