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;
int MAX=1000000007;
int n,k,a[100010],num[100010],inv[100010],ans;
int exgcd(int t1,int t2,int &x,int &y) {
	if(!t2) {
		x=1,y=0;
		return t1;
	}
	int tmp=exgcd(t2,t1%t2,y,x);
	y-=x*(t1/t2);
	return tmp;
}
int main()
{
	scanf("%d%d",&n,&k);
	for(int i=1;i<=n;++i) scanf("%d",&a[i]);
	sort(a+1,a+n+1);
	num[1]=1;inv[0]=1;
	for(int i=2;i<=n;++i) num[i]=(ll)num[i-1]*i%MAX;
	int tmp;
	exgcd(num[n],MAX,inv[n],tmp);
	inv[n]=(inv[n]%MAX+MAX)%MAX;
	for(int i=n-1;i>=1;--i) inv[i]=(((ll)(i+1)*inv[i+1])%MAX+MAX)%MAX;
	for(int i=k;i<=n;++i) {
		ans+=(ll)((ll)((ll)a[i]*num[i-1]%MAX)*inv[k-1]%MAX)*inv[i-k]%MAX;
		ans%=MAX;
	}
	for(int i=1;i+k-1<=n;++i) {
		ans-=(ll)((ll)((ll)a[i]*num[n-i]%MAX)*inv[k-1]%MAX)*inv[n-i-k+1]%MAX;
		ans=(ans%MAX+MAX)%MAX;
	}
	printf("%d",ans%MAX);
	return 0;
}