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;
int f[100001];
const int r=1000000007;
int binpower(int a,int b) {
    if (b==0)
        return 1;
    if (b%2==1)
        return ((long long)a*binpower(a,b-1))%r;
    int x=binpower(a,b/2);
    return ((long long)x*x)%r;
}
int c(int n,int m) {
    if (m>n)
        return 0;
    int x=((long long)f[m]*f[n-m])%r;
    return ((long long)f[n]*binpower(x,r-2))%r;
}
int main() {
    f[0]=1;
    int n,k;
    cin>>n>>k;
    int mas[n+1];
    for (int i=1;i<=n;i++) {
        cin>>mas[i];
        f[i]=((long long)i*f[i-1])%r;
    }
    sort(mas+1,mas+n+1);
    int ans=0;
    for (int i=1;i<=n;i++) {
        ans=(ans+(long long)mas[i]*c(i-1,k-1))%r;
        ans=(ans-(long long)mas[i]*c(n-i,k-1))%r;
        ans=(ans+r)%r;
    }
    cout<<ans<<endl;
}
