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;
typedef long long ll;
const ll MAX = 510000;
const ll mod = 1000000007;

ll fac[MAX], finv[MAX], inv[MAX];

void COMinit(){
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for(int i=2;i<MAX;i++){
        fac[i] = fac[i-1] * i % mod;
        inv[i] = mod - inv[mod%i] * (mod / i) % mod;
        finv[i] = finv[i-1] * inv[i] % mod;
    }
}
ll COM(int n, int k){
    if(n<k) return 0;
    if(n<0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n-k] % mod) % mod;
}

int main(){
    
    int n, k;
    cin >> n >> k;
    vector<ll> a(n);
    for(int i=0;i<n;i++)cin >> a[i];
    sort(a.begin(), a.end());
    
    COMinit();

    ll nyan = 0;

    for(int i=0;i<n;i++){
        nyan += -a[i] * COM(n-i-1,k-1) % mod + a[i] * COM(i,k-1) % mod;
        nyan %= mod;
    }
    cout << nyan;
}