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 int N=100005,Mod=1e9+7;
int a[N],fac[N],inv[N];
inline int qpow(int a,int b){
	int ans=1;
	while (b){
		if (b&1) ans=1ll*ans*a%Mod;
		a=1ll*a*a%Mod,b>>=1;
	}
	return ans;
}
int C(int n,int m){
	return 1ll*fac[n]*inv[m]%Mod*inv[n-m]%Mod;
}
int main (){
	int n,k;scanf ("%d%d",&n,&k);
	for (int i=1;i<=n;i++) scanf ("%d",&a[i]);
	fac[0]=1;
	for (int i=1;i<=n;i++) fac[i]=1ll*fac[i-1]*i%Mod;
	inv[n]=qpow(fac[n],Mod-2);
	for (int i=n-1;i>=0;i--) inv[i]=1ll*inv[i+1]*(i+1)%Mod;
	sort(a+1,a+n+1);int ans=0;
	for (int i=k;i<=n;i++) ans=(ans+1ll*a[i]*C(i-1,k-1))%Mod;
	for (int i=1;i<=n-k+1;i++) ans=(ans-1ll*a[i]*C(n-i,k-1)%Mod+Mod)%Mod;
	printf ("%d",ans); 
	return 0;
}
