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}.
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1100;
const int mod = 1e9 + 7;
inline void add(int &a, const int &b) {
  a += b;
  if (a >= mod) a -= mod;
}
int n, m, C[maxn][maxn], f[maxn];
bitset<maxn> a[maxn], base[maxn], M;
bool v[maxn];
char str[maxn];
int main() {
  C[0][0] = 1;
  for (int i = 1; i < maxn; i++) {
    C[i][0] = 1;
    for (int j = 1; j <= i; j++) {
      C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % mod;
    }
  }
  f[0] = 1;
  for (int i = 1; i < maxn; i++) {
    for (int j = 0; j < i; j++) {
      add(f[i], (long long)C[i - 1][j] * f[i - 1 - j] % mod);
    }
  }
  scanf("%d%d", &m, &n);
  M.set();
  for (int i = 1; i <= n; i++) {
    scanf("%s", str + 1);
    for (int j = 1; j <= m; j++) {
      if (str[j] == '1') {
        a[i][j] = 1;
      }
    }
  }
  for (int i = 1; i <= m; i++) {
    base[i] = M;
    for (int j = 1; j <= n; j++) {
      if (a[j][i])
        base[i] &= a[j];
      else
        base[i] &= (a[j] ^ M);
    }
  }
  int re = 1;
  for (int i = 1; i <= m; i++) {
    int num = 0;
    for (int j = i; j <= m; j++) {
      if (base[i][j] && !v[j]) {
        num++, v[j] = 1;
      }
    }
    re = (long long)re * f[num] % mod;
  }
  printf("%d\n", re);
  return 0;
}
