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;
const int mod = 1000000007;
const int MAXN = 100005;
LL num[MAXN],fact[MAXN];
LL my_pow(LL a, LL b){    
	LL sum=1;    
	while(b>0){        
		if(b&1){
			sum=sum*a%mod;
		}
		a=a*a%mod;        
		b>>=1;    
	}    
	return sum%mod;
}
void init(){    
	fact[0]=1;    
	for(int i=1;i<MAXN;++i){
		fact[i]=fact[i-1]*i%mod;
	}     
}
LL comb(LL a, LL b){    
	if(b>a){
		return 0;
	}   
	return fact[a]* my_pow(fact[b]*fact[a-b]%mod,mod-2)%mod;
}
int main(){   
	int n,k;   
	init();   
	cin>>n>>k;   
	for(int i=0;i<n;++i)       
		cin>>num[i];   
	LL ans=0;   
	sort(num, num+n);   
	for(int i=0;i<n-k+1;++i){       
		ans=(ans+(num[n-i-1]-num[i])%mod*comb(n-i-1,k-1)%mod)%mod;          
	}   
	cout<<(ans%mod+mod)%mod<<endl;
	return 0;
}
