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;
using ll=long long;
#define mod 1000000007
vector<ll> fact(200001);

ll power(ll a,ll r)
{
	if (r==1) return a;
	else
	{
		ll m;
		m=power(a,r/2);
		if (r%2==0) return m*m%mod;
		else return m*m%mod*a%mod;
	}	
}

void factorial()
{
	fact[0]=1;
	for (int i=1;i<200001;i++) fact[i]=fact[i-1]*i%mod;
}

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

ll cmb(ll n,ll r)
{
	if (r>n) return 0;
	if (n==r) return 1;
	return fact[n]*modinv(fact[r])%mod*modinv(fact[n-r])%mod;
}

int main()
{
	ll N,K; cin >> N >> K;
	vector<ll>A(N);
	factorial();
	ll ans=0;
	for (int i=0;i<N;i++)
	{
		cin >> A[i];
	}
	sort(A.begin(),A.end());
	for (int i=0;i<N;i++)
	{
	
	ans=(ans-A[i]*cmb(N-i-1,K-1)%mod)%mod;
	ans=(ans+A[i]*cmb(i,K-1)%mod)%mod;
	}
	cout << ans << endl;
	return 0;
}