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;
const ll zero = 0;
const ll INF = 10000000000000000; //10^16
const ll MOD = 1000000007; //10^9+7
const ll MOD2 = 998244353;


int main(void){
    vector<ll> inv(100005);
    inv.at(1) = 1;
    for(int i = 2; i < 100005; i++){
        inv.at(i) = MOD - inv.at(MOD % i) * (MOD / i) % MOD;
    }
    
    ll N, K;
    cin >> N >> K;
    vector<ll> a(N+1); for(ll i = 1; i < N+1; i++) cin >> a.at(i);
    
    sort(a.begin() + 1, a.end(), greater<ll> ());
    
    ll ans = 0;
    ll bi = 1;
    ll tmp;
    for(ll i = 1; i <= N - K + 1; i++){
        tmp = ((a.at( N - K + 2 - i ) - a.at( K + i - 1 )) + 2*MOD) % MOD;
        tmp = tmp * bi % MOD;
        ans = (ans + tmp + MOD) % MOD;
        bi = bi * ( K + i - 1 ) % MOD;
        bi = bi * inv.at( i ) % MOD;
    }
    
    printf("%lld\n", ans);
    
}