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.*;

public class Main {
    static int MAX = 510000;
    static int MOD = 1000000007;
    static long[] fac;
    static long[] finv;
    static long[] inv;
    
    public static void main(String[] args) throws Exception {
        // Your code here!
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int k = sc.nextInt();
        
        int[] array = new int[n];
        for (int i = 0; i < n; i++) {
            array[i] = sc.nextInt();
        }
        Arrays.sort(array);
        
        fac = new long[MAX];
        finv = new long[MAX];
        inv = new long[MAX];
        COMinit();
        
        long ans = 0;
        for (int i = k - 1; i < n; i++) {
            ans = (ans + (nCkMOD(i, k - 1, MOD) * array[i]) % MOD) % MOD;
        }
        
        for (int i = 0; i < n - k + 1; i++) {
            ans = (ans - (nCkMOD(n - 1 - i, k - 1, MOD) * array[i]) % MOD) % MOD;
        }
        
        if (ans < 0) {
            ans += MOD;
        }
        
        System.out.println(ans);
    }
    
    // テーブルを作る前処理
    public static void COMinit() {
        fac[0] = fac[1] = 1;
        finv[0] = finv[1] = 1;
        inv[1] = 1;
        for (int i = 2; i < MAX; i++){
            fac[i] = fac[i - 1] * i % MOD;
            inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
            finv[i] = finv[i - 1] * inv[i] % MOD;
        }
    }
    // 二項係数計算
    public static long nCkMOD(int n, int k, int MOD){
        if (n < k) return 0;
        if (n < 0 || k < 0) return 0;
        return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
    }
}
