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 ll long long
#define pb push_back
#define pii pair<int,int>
#define mp make_pair
#define fi first
#define se second
#define eb emplace_back
using namespace std;
const int mod = 1e9 + 7;
const int maxn = 2e5 + 233;
int a[maxn];
int exgcd(int a,int b,int& x,int &y){
	if(b==0){
		x = 1;y = 0;
		return a;
	}
	int d = exgcd(b,a%b,y,x);
	y -= (a/b)*x;
	return d;
}
ll inv(int i){
	int x,y;
	exgcd(i,mod,x,y);
	return (x+mod)%mod;
} 
int main()
{
	int n,k;cin>>n>>k;
	if(k==1){
		puts("0");
		return 0;
	}
	for(int i=1;i<=n;i++){
		scanf("%d",&a[i]); 
	}
	sort(a+1,a+1+n);
	ll ans = 0,cur = 1;
//	cout<<"!"<<inv(2)<<endl;
	for(int i=k;i<=n;i++){
		ans = (ans+a[i]*cur)%mod;
		ans = (ans-a[n-i+1]*cur)%mod;
		ans = (ans+mod)%mod;
		cur = (cur*i%mod*inv(i-k+1))%mod;
//		cout<<"!"<<cur<<endl;
	}
	cout << ans << endl;
	return 0;
}
