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.Arrays;
import java.util.Scanner;

public class Main {
    final static long MOD = 1000000007;

    public static void main(final String[] args) {
        final Scanner scanner = new Scanner(System.in);
        final int n = scanner.nextInt();
        final int k = scanner.nextInt();
        final long[] array = new long[n];
        for (int i = 0; i < n; i++) {
            array[i] = scanner.nextLong();
        }

        if (k == 1) {
            System.out.println(0);
            return;
        }

        Arrays.sort(array);

        final CombinationCalculator combinationCalculator = new CombinationCalculator(n + 1, (int) MOD).init();

        long sum = 0;
        for (int i = 0; i <= n - k; i++) {
            sum += array[n - 1 - i] * combinationCalculator.calc(n - 1 - i, k - 1) % MOD;
            sum %= MOD;
            sum -= array[i] * combinationCalculator.calc(n - 1 - i, k - 1) % MOD;
            sum %= MOD;
        }

        System.out.println((sum + MOD) % MOD);
    }

    private static class CombinationCalculator {
        private final int size;
        private final int mod;
        private final long[] factorials;
        private final long[] invertedElements;
        private final long[] invertedFactorials;

        CombinationCalculator(final int size, final int mod) {
            this.size = size;
            this.mod = mod;
            this.factorials = new long[size];
            this.invertedElements = new long[size];
            this.invertedFactorials = new long[size];
        }

        CombinationCalculator init() {
            factorials[0] = 1;
            factorials[1] = 1;
            invertedFactorials[0] = 1;
            invertedFactorials[1] = 1;
            invertedElements[1] = 1;
            for (int i = 2; i < size; i++) {
                factorials[i] = factorials[i - 1] * i % mod;
                invertedElements[i] = mod - invertedElements[mod % i] * (mod / i) % mod;
                invertedFactorials[i] = invertedFactorials[i - 1] * invertedElements[i] % mod;
            }
            return this;
        }

        long calc(final int n, final int k) {
            return factorials[n] * (invertedFactorials[k] * invertedFactorials[n - k] % mod) % mod;
        }
    }
}
