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;
#define all(c) (c).begin(),(c).end()
#define pb push_back 
#define mp make_pair
typedef long long ll;

ll mod=1000000007;

ll mul(ll a,ll b, ll c){
	ll ans=0;
	while(b>0){
		if(b&1) ans=(ans+a)%c;
		a=(a+a)%c;
		b>>=1;
	}
	return ans;
}

ll ex2(ll a,ll b,ll c){
	ll ans=1;
	while(b>0){
		if(b&1) ans=mul(ans,a,c);
		a=mul(a,a,c);
		b>>=1;
	}
	return ans;
}

int main(){
	int n,k; cin>>n>>k;
	vector<ll> a(n);
	
	for(int i=0;i<n;i++) cin>>a[i];
			
	sort(all(a));
	
	ll c[n+5];
	
	c[k-1]=1ll;
	
	for(int i=k; i<=n-1; i++){
		c[i]=mul(c[i-1],(ll) i, mod);
		c[i]=mul(c[i],ex2(i-k+1, mod-2, mod), mod);
	}
	
	/*for(int i=k-1;i<=n-1;i++) cout<<c[i]<<" ";
	cout<<endl;*/
	
	ll ans=0;
	
	for(int i=0; i<=n-k; i++){
		ans+=mul(a[n-1-i] - a[i], c[n-1-i], mod);
		ans%=mod;
		if(ans<0) ans+=mod;
	}
	
	cout<<ans<<endl;
			
	return 0;
}
