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;

const long long mod=1000000007;
long long n, k, a[100001], ans=0, facto[100004];

long long modpow(long long x, long long y){
	long long ret=1;
	while(y>0){
		if (y&1){
			ret*=x;
			ret%=mod;
		}
		x*=x;
		x%=mod;
		y>>=1;
	}
	return ret;
}

long long modinv(long long x){
	return modpow(x,mod-2);
}

long long bino(long long aa, long long bb){
	return (((facto[aa]*modinv(facto[aa-bb]))%mod)*modinv(facto[bb]))%mod;
}

void init(){
	facto[0]=1;
	for (int i=1;i<100004;i++){
		facto[i]=facto[i-1]*i;
		facto[i]%=mod;
	}
}

int main(){
	init();
	cin >> n >> k;
	for (int i=0;i<n;i++) cin >> a[i];
	sort(a,a+n);
	for (int i=k-1;i<n;i++){
		ans+=a[i]*bino(i,k-1);
		ans%=mod;
	}
//	cout << ans << "\n";
	for (int i=0;i<=n-k;i++){
		ans-=a[i]*bino(n-i-1,k-1);
		ans%=mod;
		if (ans<0) ans+=mod;
	}
	cout << ans << "\n";
}