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
import java.util.*;
import static java.lang.Math.*;
import java.math.BigInteger;

public class Main {
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		
		// 入力
		long MOD = 1000000007;
		int n = sc.nextInt();
		int k = sc.nextInt();
		int[] a = new int[n];
		for(int i = 0; i < n; i++){
		    a[i] = sc.nextInt();
		}
		Arrays.sort(a);
		
		// 計算
		long result = 0;
		long maxSum = 0;
		long minSum = 0;
		Combination c = new Combination(n);
		for(int i = 0; i < n; i++){
		    long now = a[i];
		    maxSum += now*c.combination(i, k-1)%MOD;
		    minSum += now*c.combination(n-(i+1), k-1)%MOD;
		}
		result = maxSum + MOD - minSum;
		result %= MOD;
		
		// 出力
		System.out.println(result);
	}
}

class Combination {
    static final int MOD = 1000000007;
    long[] fact;
    long[] invFact;
    long[] inv;
    
    public Combination (int n) {
        fact = new long[n+1];
        invFact = new long[n+1];
        inv = new long[n+1];
        fact[0] = 1;
        fact[1] = 1;
        invFact[0] = 1;
        invFact[1] = 1;
        inv[1] = 1;
        for (int i = 2; i < n+1; i++) {
            fact[i] = fact[i - 1] * i % MOD;
            inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;
            invFact[i] = invFact[i - 1] * inv[i] % MOD;
        }
    }
    
    public long combination(int n, int k) {
        if (n < k || n < 0 || k < 0) {
            return 0;
        }
        return fact[n] * (invFact[k] * invFact[n - k] % MOD) % MOD;
    }
}


