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 ll long long int
#define MOD 1000000007
ll f[100001];
ll power(ll a, ll b)
{
 ll x=1,y=a;
 while(b > 0)
 	{
 		if(b%2 == 1)
 	{
 		x=(x*y);
 		if(x>MOD) x%=MOD;
 	}
 	y = (y*y);
 	if(y>MOD) y%=MOD;
 	b /= 2;
 	}
 return x;
}
ll InverseEuler(ll n)
{
 return power(n,MOD-2);
}
ll C(ll n, ll r)
{
 return (f[n]*((InverseEuler(f[r]) * InverseEuler(f[n-r])) % MOD)) % MOD;
}
int main()
{  ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    ll n,k;
    f[0] = 1;
	for(int i = 1 ; i <= 100000 ; i++)
		f[i] = (f[i-1]*i)%MOD;
    cin>>n>>k;
    ll a[n+1];
    for(int x=1;x<=n;x++)
        cin>>a[x];
    sort(a+1,a+n+1);
    ll g=0;
    for(int x=1;x<=n;x++)
    {   if(n-x>=k-1)
        {g+=((a[x])*C(n-x,k-1))%MOD;
        g%=MOD;
        }
    }
    ll j=0;
    for(int x=k; x<=n; x++){
        j+=(a[x]*C(x-1,k-1))%MOD;
        j%=MOD;
    }
    cout<<(MOD+(j-g))%MOD;
}
