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 ld long double
#define pii pair<int,int>
#define fi first
#define se second
#define mex 1000005
//vector<int> adj[mex];
int mod=1e9+7;
long long fact[100005];

long long power(long long a,long long b){
	long long res=1;
	while(b){
		if(b&1)
		 res=(res*a)%mod;
		b=b/2;
		a=(a*a)%mod;
	}
	return res;
}
 
long long comb(long long n,long long k){
	if(k>n || k<0)
	 return 0;
	return ((fact[n]*power(fact[k],mod-2))%mod*power(fact[n-k],mod-2))%mod;
}
 
int main(){
	int n,k;
	cin>>n>>k;
	vector<long long int> v(n);
	for(int i=0;i<n;i++)
	 cin>>v[i];
	sort(v.begin(),v.end());
	fact[0]=1;
	for(int i=1;i<=n;i++)
	 fact[i]=(fact[i-1]*i)%mod;
	long long sum1=0,sum=0;
	for(int i=k-1;i<n;i++){
		long long tx=(v[i]*comb(i,k-1))%mod;
		long long ti=(v[n-i-1]*comb(i,k-1))%mod; 
		sum=(sum+tx)%mod;
		sum1=(sum1+ti)%mod;
	}
	cout<<(sum-sum1)%mod;

}


