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;

long inv(int a, int m) {
    long i = 1, j = 0;
    for(int b = m; a > 1; swap(i, j), swap(a, b)) {
        j = (j + b / a * (m - i)) % m;
        b %= a;
    }
    return i;
}
long f[111111] = {1};
int a[111111];
int main() {
    int n, k, p = 1e9 + 7;
    cin >> n >> k;
    for(int i = 0; i < n; i++)
        scanf("%d", a + i);
    sort(a, a + n);
    for(int i = 0; i < n; i++)
        f[i + 1] = f[i] * (i + 1) % p;
    long r = 0;
    for(int i = 0; i < n; i++) {
        if(i >= k - 1)
            (r += f[i] * a[i] % p * inv(f[k - 1] * f[i - k + 1] % p, p)) %= p;
        if(n - i >= k)
            (r += (p - f[n - i - 1]) * a[i] % p *
                  inv(f[k - 1] * f[n - i - k] % p, p)) %= p;
    }
    cout << r << endl;
    return 0;
}
