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 <iostream>
#include <algorithm>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <string>
#include <iomanip>
using namespace std;
long long mod=1e9+7;
long long modinv(long long a,long long m){
	long long b=m,u=1,v=0;
	while(b){
		long long t=a/b;
		a-=t*b;
		swap(a,b);
		u-=t*v;
		swap(u,v);
	}
	(u+=mod)%=mod;
	return u;
}
int main(){
	long long n,k;
	long long a[100000];
	cin>>n>>k;
	for(int i=0;i<n;i++){
		cin>>a[i];
	}
	sort(a,a+n);
	long long cul[100001];
	cul[k-1]=1;
	for(int i=k;i<=n-1;i++){
		cul[i]=cul[i-1];
		(cul[i]*=i)%=mod;
		(cul[i]*=modinv(i-(k-1),mod))%=mod;
	}
	long long ans=0;
	for(int i=0;i<=n-k;i++){
		ans+=(mod-cul[n-i-1]*a[i]);
		ans%=mod;
	}
	for(int i=k-1;i<n;i++){
		ans+=cul[i]*a[i];
		ans%=mod;
	}
	cout<<ans<<endl;
	return 0;
}