You are given an integer m.

Let M = 2m - 1.

You are also given a set of n integers denoted as the set T. The integers will be provided in base 2 as n binary strings of length m.

A set of integers S is called "good" if the following hold. 

  1. If <image>, then <image>. 
  2. If <image>, then <image>
  3. <image>
  4. All elements of S are less than or equal to M. 



Here, <image> and <image> refer to the bitwise XOR and bitwise AND operators, respectively.

Count the number of good sets S, modulo 109 + 7.

Input

The first line will contain two integers m and n (1 ≤ m ≤ 1 000, 1 ≤ n ≤ min(2m, 50)).

The next n lines will contain the elements of T. Each line will contain exactly m zeros and ones. Elements of T will be distinct.

Output

Print a single integer, the number of good sets modulo 109 + 7. 

Examples

Input

5 3
11010
00101
11000


Output

4


Input

30 2
010101010101010010101010101010
110110110110110011011011011011


Output

860616440

Note

An example of a valid set S is {00000, 00101, 00010, 00111, 11000, 11010, 11101, 11111}.
MOD = 10**9 + 7
m, N = map(int, input().split())

binom = [[1] + [0 for i in range(m)] for j in range(m + 1)]
for n in range(1, m + 1):
    for k in range(1, n + 1):
        binom[n][k] = (binom[n - 1][k] + binom[n - 1][k - 1]) % MOD

bell = [0 for n in range(m + 1)]
bell[0] = bell[1] = 1
for n in range(1, m):
    for k in range(n + 1):
        bell[n + 1] += bell[k] * binom[n][k]
        bell[n + 1] %= MOD
#print(bell)

bags = [0 for i in range(m)]
for it in range(N):
    for i, z in enumerate(input()):
        if z == '1':
            bags[i] |= (1 << it)
difs = set(bags)
sol = 1
for mask in difs:
    sol = sol * bell[bags.count(mask)] % MOD
print(sol)
