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>
#define mod 1000000007
using namespace std;
int n,k,a[100005];
int fac[100005],ifac[100005];
int ans1,ans2;

int ksm(int x,int y){
	int res=1;
	while(y){
		if(y&1) res=1ll*res*x%mod;
		x=1ll*x*x%mod,y/=2;
	}
	return res;
}

int C(int x,int y){
	int res=1ll*fac[x]*ifac[y]%mod*ifac[x-y]%mod;
	return res;
}

int main(){
	scanf("%d %d",&n,&k);
	for(int i=1;i<=n;i++) scanf("%d",&a[i]);
	sort(a+1,a+n+1);
	for(int i=1;i<=n;i++) if(a[i]<0) a[i]+=mod;
	fac[0]=1;for(int i=1;i<=n;i++) fac[i]=1ll*fac[i-1]*i%mod;
	ifac[n]=ksm(fac[n],mod-2);
	for(int i=n-1;i>=0;i--) ifac[i]=1ll*ifac[i+1]*(i+1)%mod;
	for(int i=k;i<=n;i++){
		ans1+=1ll*a[i]*C(i-1,k-1)%mod;
		if(ans1>=mod) ans1-=mod;
	}
	for(int i=1;i<=n-k+1;i++){
		ans2+=1ll*a[i]*C(n-i,k-1)%mod;
		if(ans2>=mod) ans2-=mod;
	}
	printf("%d\n",(ans1-ans2+mod)%mod);

	return 0; 
}