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;

int main()
{
    ll i,j;
    ll n,k;
    cin >> n >> k;
    vector<ll> a(n);
    for(i = 0;i < n;++i){
        cin >> a.at(i);
    }
    ll ans = 0;
    sort(a.begin(), a.end());
    
    int max_number = n;
    int MOD = 1e9+7;
    vector<ll> fnc(max_number+1,1);
    vector<ll> inv(max_number+1,1);
    for(i = 2;i <= max_number;++i){
        fnc.at(i) = fnc.at(i-1)*i%MOD;
        inv.at(i) = MOD - inv.at(MOD%i)*(MOD/i)%MOD;
    }
    vector<ll> finv(max_number+1,1);
    for(i = 2;i <= max_number;++i){
        finv.at(i) = finv.at(i-1)*inv.at(i)%MOD;
    }
    for(i = k-1;i < n;++i){
        ans += a.at(i)*(fnc.at(i)*(finv.at(i-k+1)*finv.at(k-1)%MOD)%MOD)%MOD;
        ans %= MOD;
    }
    for(i = 0;i < n-k+1;++i){
        ans += (MOD - a.at(i)*(fnc.at(n-1-i)*(finv.at(n-k-i)*finv.at(k-1)%MOD)%MOD)%MOD);
        ans %= MOD;
    }
    cout << ans << endl;
    return 0;
}