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 ll long long
#define vi vector<int>
#define pii pair<int,int>
#define vii vector<pair<int,int>>

ll MOD = 1e9+7;
ll inv(ll a, ll b) {
	return 1 < a ? b - inv(b % a, a) * b / a : 1;
}

ll fac[100001];
ll C(ll n, ll m){
    return (((fac[n]*inv(fac[m],MOD))%MOD)*(inv(fac[n-m], MOD)))%MOD;
}

int main(){
    ios::sync_with_stdio(0);
    int n, k;
    ll ans = 0;
    cin >> n >> k;
    ll a[n], b[n], t = n-1;
    for (int i = 0 ; i < n ; ++i){
        cin >> a[i];
        b[i] = a[i];
    }
    sort(a, a+n);
    fac[0] = 1;
    for (int i = 1 ; i < 100001 ; ++i)
        fac[i] = (fac[i-1]*i)%MOD;

	for(int i = 0;i<n;i++){
		if(i>=k-1)ans+=a[i]*C(i,k-1),ans%=MOD;
		if(n-i-1>=k-1)ans-=a[i]*C(n-i-1,k-1),(ans+=MOD)%=MOD;
	}
	cout << ans;
    return 0;
}
