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 FOR(i,a,b) for(int i=(a);i<(b);++i)
#define rep(i,n)   FOR(i,0,n)
#define pb emplace_back
typedef long long ll;
typedef pair<int,int> pint;

const int MAX_N=100001;
const int mod=1000000007;
ll fact[MAX_N+1];
ll inv[MAX_N+1],finv[MAX_N+1];
int inv_init(ll n,ll p){
    inv[1]=1;
    FOR(i,2,n+1) inv[i]=p-(p/i)*inv[p%i]%p;
    fact[0]=finv[0]=1;
    FOR(i,1,n+1){
        fact[i]=fact[i-1]*i%p;
        finv[i]=finv[i-1]*inv[i]%p;
    }
    return 0;
}
ll comb_mod(ll n,ll k,ll p){
    if(k<0||n<k) return 0;
    return finv[k]*finv[n-k]%p*fact[n]%p;
}
int n,k;
ll a[100001];
int main(){
    cin>>n>>k;
    rep(i,n) cin>>a[i];
    sort(a,a+n);
    inv_init(n+1,mod);
    ll r=0,l=0;
    rep(i,n){
        r+=comb_mod(i,k-1,mod)*a[i];
        l+=comb_mod(n-i-1,k-1,mod)*a[i];
        r%=mod,l%=mod;
    }
    cout<<(r+mod-l)%mod<<endl;
    return 0;
}