p03406 AtCoder Regular Contest 093 - Dark Horse
There are 2^N players, numbered 1, 2, ..., 2^N. They decided to hold a tournament.

The tournament proceeds as follows:

* Choose a permutation of 1, 2, ..., 2^N: p_1, p_2, ..., p_{2^N}.
* The players stand in a row in the order of Player p_1, Player p_2, ..., Player p_{2^N}.
* Repeat the following until there is only one player remaining in the row:
* Play the following matches: the first player in the row versus the second player in the row, the third player versus the fourth player, and so on. The players who lose leave the row. The players who win stand in a row again, preserving the relative order of the players.
* The last player who remains in the row is the champion.



It is known that, the result of the match between two players can be written as follows, using M integers A_1, A_2, ..., A_M given as input:

* When y = A_i for some i, the winner of the match between Player 1 and Player y (2 \leq y \leq 2^N) will be Player y.
* When y \neq A_i for every i, the winner of the match between Player 1 and Player y (2 \leq y \leq 2^N) will be Player 1.
* When 2 \leq x < y \leq 2^N, the winner of the match between Player x and Player y will be Player x.



The champion of this tournament depends only on the permutation p_1, p_2, ..., p_{2^N} chosen at the beginning. Find the number of permutation p_1, p_2, ..., p_{2^N} chosen at the beginning of the tournament that would result in Player 1 becoming the champion, modulo 10^9 + 7.

Constraints

* 1 \leq N \leq 16
* 0 \leq M \leq 16
* 2 \leq A_i \leq 2^N (1 \leq i \leq M)
* A_i < A_{i + 1} (1 \leq i < M)

Input

Input is given from Standard Input in the following format:


N M
A_1 A_2 ... A_M


Output

Print the answer.

Examples

Input

2 1
3


Output

8


Input

4 3
2 4 6


Output

0


Input

3 0


Output

40320


Input

3 3
3 4 7


Output

2688


Input

16 16
5489 5490 5491 5492 5493 5494 5495 5497 18993 18995 18997 18999 19000 19001 19002 19003


Output

816646464
```python
import sys
input = sys.stdin.readline

def main():
    def com(com_n, com_r):
        return fac[com_n] * inv[com_r] * inv[com_n - com_r] % md

    md = 10 ** 9 + 7
    n, m = map(int, input().split())
    aa = list(map(lambda x: int(x) - 1, input().split()))
    n2 = 2 ** n

    # combinationの準備
    n_max = n2
    fac = [1]
    inv = [1] * (n_max + 1)
    k_fac_inv = 1
    for i in range(1, n_max + 1):
        k_fac_inv = k_fac_inv * i % md
        fac.append(k_fac_inv)
    k_fac_inv = pow(k_fac_inv, md - 2, md)
    for i in range(n_max, 1, -1):
        inv[i] = k_fac_inv
        k_fac_inv = k_fac_inv * i % md

    dp = [[0] * n2 for _ in range(m + 1)]
    dp[0][0] = 1
    for i in range(m):
        a = aa[m - 1 - i]
        for b in range(n2):
            pre = dp[i][b]
            if pre == 0: continue
            dp[i + 1][b] = (dp[i + 1][b] + pre) % md
            k = 1
            for _ in range(n):
                if b & k == 0 and n2 - a - b >= k:
                    nb = b | k
                    dp[i + 1][nb] = (dp[i + 1][nb] + pre * com(n2 - 1 - a - b, k - 1)) % md
                k <<= 1
    for b in range(n2):
        k = 1
        p = n2 - 1 - b
        for _ in range(n):
            if b & k == 0:
                dp[m][b] = dp[m][b] * com(p, k) % md
                p -= k
            k <<= 1
    ans = 0
    for b in range(n2):
        coff = -1 if bin(b).count("1") % 2 else 1
        ans = ans + coff * dp[m][b]
    k = 1
    for _ in range(n - 1):
        k <<= 1
        ans = ans * fac[k] % md
    ans = ans * n2 % md
    print(ans)

main()

```