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 <iostream>
#include <algorithm>
typedef long long ll;
using namespace std;
ll C[100005];
ll A[100005];
ll inv(ll K, ll M) {
  ll A = M - 2;
  ll ret = 1;
  ll D = K;
  while (A > 0) {
    if (A % 2 == 1) {
      ret = (ret * D) % M;
    }
    D = (D * D) % M;
    A /= 2;
  }
  return ret;
}
int main() {
  ios::sync_with_stdio(false);
  int N, K;
  cin >> N >> K;
  for (int i=0; i<N; i++) {
    cin >> A[i];
  }
  sort(A, A+N);
  ll S = 0;
  ll M = 1000000007LL;
  C[K-1] = 1;
  ll I = inv(N+1-K, M);
  for (ll i=K; i<=N; i++) {
    C[i] = (C[i-1] * i) % M;
    C[i] = (C[i] * inv(i-(K-1), M)) % M;
  }
  for (ll i=0; i<N; i++) {
    ll m = (C[i] - C[N-1-i]) % M;
    m = (m + M) % M;
    ll d = (m * A[i]) % M;
    d = (d + M) % M;
    S = (S + d) % M;
  }
  cout << S << endl;
}