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 F first
#define S second
using namespace std;
typedef pair<int,int> pii;
typedef long long ll;
int n,k; ll arr[101010],fac[101010],ans;
const ll MOD=1000000007;
ll factorial(int p)
{
	if(p==0) return fac[0]=1;
	return fac[p]=(factorial(p-1)*p)%MOD;
} ll pow(ll p,int q)
{
	if(q==0) return 1;
	if(q==1) return p;
	ll ret=pow(p,q/2);
	if(q%2) return (((ret*ret)%MOD)*p)%MOD;
	return (ret*ret)%MOD;
}
ll combi(int p,int q)
{
	if(p<q) return 0;
	ll imsi=(fac[q]*fac[p-q])%MOD;
	imsi=pow(imsi,1000000005); 
	return (fac[p]*imsi)%MOD;
}
int main()
{
	cin>>n>>k; factorial(n);
	for(int i=1;i<=n;i++)
	{
		scanf("%lld",arr+i);
	} sort(arr+1,arr+n+1);
	for(int i=1;i<=n;i++)
	{
		ans+=arr[i]*(combi(i-1,k-1)-combi(n-i,k-1)); 
		ans=(ans+MOD*10)%MOD;
	} printf("%d",ans);
}