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 <cstdio>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <vector>
 
using namespace std;
#define PB push_back
#define MP make_pair
#define REP(i,n) for(int i=0;i<(n);i++)
#define ll long long
const int mod=1e9+7;
ll n,k,a[100100];
ll F[1<<17],I[1<<17];
ll C(ll a,ll b){return F[a]*I[b]%mod*I[a-b]%mod;}
ll P(ll a,ll b){return b?P(a*a%mod,b/2)*(b%2?a:1)%mod:1;}
int main(){
    int n,k,a[100100];
    cin>>n>>k;
    REP(i,n)cin>>a[i];

    F[0]=1;
    for(int i=1;i<=n;i++)F[i]=F[i-1]*i%mod;

    I[n]=P(F[n],mod-2);
    for(int i=n-1;i>=0;i--)I[i]=I[i+1]*(i+1)%mod;

    sort(a,a+n);

    ll ans=0;
    for(int i=k-1;i<n;i++)
    {
        ans=(ans+a[i]*C(i,k-1))%mod;
        ans=(ans-a[n-i-1]*C(i,k-1))%mod;
    }
    cout<<ans<<endl;
}