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 int long long
#define f first
#define s second
using namespace std;
const int mod=1e9+7;
const int N=1e5+9;
int fa[N],iv[N];
int f(int a,int n=mod-2){
	if(n==0)return 1;
	if(n==1)return a;
	int x=f(a,n/2);
	x=(x*x)%mod;
	if(n%2)x=(a*x)%mod;
	return x;
}
int c(int n,int k){
	return ((fa[n]*iv[k])%mod*iv[n-k])%mod;
}
int32_t main(){
	// freopen("input.txt", "r", stdin);
	// freopen("output.txt", "w", stdout);
	ios::sync_with_stdio(0),cin.tie(0);

	int n,k;
	cin >> n >> k;
	int a[n];
	fa[0]=1;iv[0]=1;
	for(int i=1;i<N;i++){
		fa[i]=(fa[i-1]*i)%mod;
		iv[i]=f(fa[i]);
	}
	for(int i=0;i<n;i++)
		cin >> a[i];
	int sum=0;
	sort(a,a+n);
	for(int i=0;i<=n-k;i++){
		sum=(sum+mod-(a[i]*c(n-1-i,k-1))%mod)%mod;
	}
	for(int i=k-1;i<n;i++){
		sum=(sum+(a[i]*c(i,k-1))%mod)%mod;
	}
	cout << sum;
}
