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;

#define endl '\n'

long long const N = 1e5 + 1, M = 1e9 + 7;
long long fn[N], fd[N], iv[N];

void initnck(){
    iv[1] = fn[1] = fd[1] = fn[0] = fd[0] = 1;
    for (long long i=2; i<N; i++){
        iv[i] = (M - (M/i) * iv[M%i] % M) % M;
        fn[i] = fn[i-1]*i%M;
        fd[i] = fd[i-1]*iv[i]%M;
    }
}

long long nck(long long n, long long k){
    return fn[n] * fd[n-k] % M * fd[k] % M;
}

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(0);

    initnck();
    long long n,k, a[N]; cin >> n >> k;

    for (int i=1; i<=n; i++) cin >> a[i];
    sort(a+1, a+n+1);

    long long maxs = 0, mins = 0;
    for (int i=1; i<=n; i++){
        if (i>=k) maxs = (maxs + a[i] * nck(i-1, k-1) + M)%M;
        if (i<=n-k+1) mins = (mins + a[i] * nck(n-i, k-1) + M)%M;
    }
    cout << (maxs - mins + M)%M << endl;
}