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>

using namespace std;

const int Z = 1e9+7;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(NULL);

    int n, k;
    cin >> n >> k;
    int a[n];
    for (int i = 0; i < n; i++) {
        cin >> a[i];
    }
    sort(a, a+n);

    int inv[n+1];
    for (int i = 1; i <= n; i++) {
        inv[i] = i == 1 ? 1 : 1ll * (Z - Z / i) * inv[Z%i] % Z;
    }

    int fac[n+1], ifac[n+1];
    for (int i = 0; i <= n; i++) {
        fac[i] = i == 0 ? 1 : 1ll * fac[i-1] * i % Z;
        ifac[i] = i == 0 ? 1 : 1ll * ifac[i-1] * inv[i] % Z;
    }

    int ans = 0;
    for (int i = 0; i <= n - k; i++) {
        ans = ((ans - 1ll * fac[n-i-1] * ifac[k-1] % Z * ifac[n-i-k] % Z * a[i]) % Z + Z) % Z;
    }
    for (int i = k - 1; i < n; i++) {
        ans = ((ans + 1ll * fac[i] * ifac[k-1] % Z * ifac[i+1-k] % Z * a[i]) % Z + Z) % Z;
    }

    cout << ans << endl;

    return 0;
}
