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 mod = 1e9+7;

const ll MAX = 1e7;
ll fac[MAX], finv[MAX], inv[MAX];

//前処理
void COMinit (){
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for(ll i = 2; i < MAX; i++){
        fac[i] = fac[i-1]*i%mod;
        inv[i] = mod-inv[mod%i]*(mod/i)%mod;
        finv[i] = finv[i-1]*inv[i]%mod;
    }
}
//二項係数
ll COM(ll n, ll k){
    if(n < k) return 0;
    if(n < 0 || k < 0) return 0;
    return (fac[n]*(finv[k]*finv[n-k]%mod))%mod;
}

int main(){
    ll n,k;
    cin >> n >> k;
    vector<ll> a(n);
    COMinit();
    for(ll i = 0; i < n; i++) cin >> a[i];
    sort(a.begin(),a.end());
    ll ans = 0;
    if(k == 1){
        cout << 0 << endl;
        return 0;
    }
    for(ll i = 0; i <= n-k; i++){
        ans += COM(n-i-1,k-1)*((a[n-i-1]-a[i])%mod);
        ans %= mod;
    }
    cout << (ans+mod)%mod << endl;
    return 0;
}