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>
#define lint long long
#define MOD 1000000007


using namespace std;

int n,k;

lint fat[100010];


lint poww(lint b, lint e){
	lint aux = 1;
	for( ;e ;e/=2){
		if(e%2)
			aux *= b;
		aux %= MOD;
		b *= b;
		b %= MOD;
	}

	return aux;
}

lint ncr(lint n, lint r){
    if(r < 0 || n < 0) return 0;
    if(n < r) return 0;
    lint a = fat[n];
    a = (a * poww(fat[r],MOD-2)) % MOD;
    a = (a * poww(fat[n-r],MOD-2)) % MOD;
    return a;
}

int main(){


	ios_base::sync_with_stdio(false);
	cin.tie(NULL); cout.tie(NULL);
	
	fat[0] = 1;
	for(int i =1;i<=100008;i++){
		fat[i]=(i*fat[i-1])%MOD;
	}
	
	cin>>n>>k;
	
	lint ans = 0;

	vector<lint> v(n);
	for(int i=0;i<n;i++) cin>>v[i];
	sort(v.begin(), v.end());
	
	for(int i=0;i<n;i++){

		ans += ncr(i,k-1)*v[i];
		ans %= MOD;
		ans -= ncr(n-1-i,k-1)*v[i];
		ans %= MOD;
	}
	
	if(ans<0) ans += MOD;
	cout<<ans<<"\n";


		

	return 0;	
}
