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 <vector>
#include <algorithm>
using namespace std;

//  a^b % m
long long powmod(long long a, long long b, long long m)
{
    long long r = 1;
    for (; b>0; b>>=1, a=a*a%m)
        if (b&1)
            r = r*a%m;
    return r;
}

int main()
{
    int N, K;
    cin>>N>>K;
    vector<long long> A(N);
    for (long long &a: A)
        cin>>a;

    long long M = 1000000007LL;
    vector<long long> F(N+1);
    F[0] = 1;
    for (int i=1; i<=N; i++)
        F[i] = F[i-1]*i%M;

    auto C = [&](long long a, long long b)
    {
        return F[a]*powmod(F[b], M-2, M)%M*powmod(F[a-b], M-2, M)%M;
    };

    sort(A.begin(), A.end());
    long long ans = 0;
    for (int i=0; i<N; i++)
    {
        if (i>=K-1)
            ans += A[i]*C(i, K-1)%M;
        if (N-i-1>=K-1)
            ans -= A[i]*C(N-i-1, K-1)%M;
        ans = (ans+M)%M;
    }
    cout<<ans<<endl;
}
