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 <iostream>
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll n,k,a,ans1,ans2,MOD=1e9+7,bik[200001];
vector<ll> v;
ll f(ll x,ll y){
    if(y==1)return x;
    if(y%2==0)return f(x,y/2)*f(x,y/2)%MOD;
    return f(x,y-1)*x%MOD;
}
ll comb(ll n,ll r){
    return bik[n]*f(bik[r],MOD-2)%MOD*f(bik[n-r],MOD-2)%MOD;
}
int main(void){
    bik[0]=1;
    for(ll i=0;i<200000;i++){
        bik[i+1]=bik[i]*(i+1);
        bik[i+1]%=MOD;
    }
    cin>>n>>k;
    for(int i=0;i<n;i++){
        cin>>a;
        v.push_back(a);
    }
    sort(v.begin(),v.end());
    for(int i=0;i<n-k+1;i++){
        ans1+=comb(n-i-1,k-1)*v[i];
        ans1%=MOD;
    }
    for(int i=n-1;i>=k-1;i--){
        ans2+=comb(i,k-1)*v[i];
        ans2%=MOD;
    }
    ans2-=ans1;
    ans2%=MOD;
    ans2=(ans2+MOD)%MOD;
    cout<<ans2<<endl;
}
