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;
const int MOD = 1e9+7;
long long ipow(long long a, int b)
{
    if(b==1) return a;
    long long x = ipow(a, b/2);
    x = x*x%MOD;
    if(b%2==1) x = x*a%MOD;
    return x;
}

long long fact[101010];
long long invfact[101010];
long long arr[101010];
long long C(int a, int b)
{
    if(a<b) return 0;
    return fact[a]*invfact[b]%MOD*invfact[a-b]%MOD;
}
int main()
{
    fact[0] = 1;
    for(int i=1; i<101010; ++i)
        fact[i] = i*fact[i-1]%MOD;    
    for(int i=0; i<101010; ++i)
        invfact[i] = ipow(fact[i], MOD-2);
    int N, K; scanf("%d%d", &N, &K);
    for(int i=0; i<N; ++i) scanf("%lld", arr+i);
    sort(arr, arr+N);
    long long ans = 0;
    for(int i=0; i<N; ++i)
    {
        ans += arr[i]*C(i, K-1);
        ans -= arr[i]*C(N-i-1, K-1);
        ans %= MOD;
    }
    ans += MOD; ans %= MOD;
    cout << ans << endl;
    
}