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;
const int mod = (int)1e9+7;

ll power(ll a, ll b){
	ll ans = 1;
	while(b){
		if(b&1){
			ans = (ans * a)%mod;
		}
		b = b>>1;
		a = (a * a)%mod;	
	}
	return ans;
}

ll inv(ll a){
	return power(a,mod-2);
}

vector<ll> f(100005);

ll C(ll n, ll r){
	if(n<r)
		return 0;
	return  (f[n] * inv(f[r] * f[n-r] %mod) )%mod;	
}

int main(){
	ios::sync_with_stdio(false);
	cin.tie(NULL);
	
	ll n,k;
	cin>>n>>k;
	ll a[n+1];
	for(ll i=1;i<=n;i++){
		cin>>a[i];
	}
	sort(a+1,a+n+1);
	
	f[0] = 1ll; 
	for(ll i=1;i<=100004;i++){
		f[i] = (f[i-1] * i) %mod;
	}

	ll ans = 0;
	
	for(int i=1;i<=n;i++){
		ans = (ans + (a[i] * C(i-1,k-1))%mod )%mod;
		ans = (ans - (a[i] * C(n-i,k-1))%mod + mod )%mod;
	}
	cout<<ans;
	
	return 0;
}
	
	
	
	
	
	
	
	
	
	
	























