Permutation p is an ordered set of integers p1, p2, ..., pn, consisting of n distinct positive integers, each of them doesn't exceed n. We'll denote the i-th element of permutation p as pi. We'll call number n the size or the length of permutation p1, p2, ..., pn.

We'll call position i (1 ≤ i ≤ n) in permutation p1, p2, ..., pn good, if |p[i] - i| = 1. Count the number of permutations of size n with exactly k good positions. Print the answer modulo 1000000007 (109 + 7).

Input

The single line contains two space-separated integers n and k (1 ≤ n ≤ 1000, 0 ≤ k ≤ n).

Output

Print the number of permutations of length n with exactly k good positions modulo 1000000007 (109 + 7).

Examples

Input

1 0


Output

1


Input

2 1


Output

0


Input

3 2


Output

4


Input

4 1


Output

6


Input

7 4


Output

328

Note

The only permutation of size 1 has 0 good positions.

Permutation (1, 2) has 0 good positions, and permutation (2, 1) has 2 positions.

Permutations of size 3:

  1. (1, 2, 3) — 0 positions
  2. <image> — 2 positions
  3. <image> — 2 positions
  4. <image> — 2 positions
  5. <image> — 2 positions
  6. (3, 2, 1) — 0 positions

import java.util.Arrays;
import java.util.Scanner;

public class E {
    static int n;
    static int[][][] DP;
    static int mod = (int) 1e9 + 7;

    public static int get(int index, int k, int used) {
        if (k < 0)
            return 0;
        if (index == n)
            if (k == 0)
                return 1;
            else
                return 0;
        if (DP[index][k][used] != -1)
            return DP[index][k][used];
        int ans = 0;
        // none
        ans = (ans + get(index + 1, k, (used << 1) & 3)) % mod;
        // i - 1
        if ((used & 2) == 0)
            ans = (ans + get(index + 1, k - 1, (used << 1) & 3)) % mod;
        // i+1
        if (index != n - 1)
            ans = (ans + get(index + 1, k - 1, ((used << 1) & 3) + 1)) % mod;
        return DP[index][k][used] = ans;

    }

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        n = in.nextInt();
        int k = in.nextInt();
        long[] F = new long[n + 1];
        F[0] = 1;
        for (int i = 1; i <= n; i++)
            F[i] = (F[i - 1] * i) % mod;
        long[][] C = new long[n + 1][n + 1];
        for (int i = 0; i <= n; i++)
            for (int j = 0; j <= i; j++)
                if (j == 0 || i == j)
                    C[i][j] = 1;
                else
                    C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % mod;
        DP = new int[n][n + 1][4];
        for (int i = 0; i < n; i++)
            for (int j = 0; j <= n; j++)
                Arrays.fill(DP[i][j], -1);
        long[] A = new long[n + 1];
        for (int i = 0; i <= n; i++)
            A[i] = (get(0, i, 2) * F[n - i]) % mod;
        for (int i = n - 1; i >= 0; i--)
            for (int j = i + 1; j <= n; j++)
                A[i] = ((A[i] - (C[j][i] * A[j]) % mod) + mod) % mod;
        System.out.println(A[k]);
    }
}
