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;
ll gya(ll n,ll p,ll m){
	ll b=1;
	while (p>0) {
		if(p%2)b=(b*n)%m;
		n=(n*n)%m;
		p/=2;
	}
	return b;
}
int main(){
	ll n,k;
	cin>>n>>k;
	ll a[214514];
	for(int i=0;i<n;i++)cin>>a[i];
	sort(a,a+n);
	ll ans=0,mod=1e9+7;
	ll x=1,y=1;
	for(int i=1;i<k;i++){
		x*=n+1-i;
		x%=mod;
		y*=i;
		y%=mod;
	}
	x=x*gya(y,mod-2,mod);
	x%=mod;
	//cout<<x<<endl;
	for(int i=n;i>=k;i--){
		x*=i-k+1;
		x%=mod;
		x*=gya(i,mod-2,mod);
		x%=mod;
		//cout<<x<<endl;
		ans+=(3*mod+a[i-1]-a[n-i])*x;
		ans%=mod;
		//cout<<ans<<endl;
	}
	cout<<ans<<endl;
	return 0;
}
