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>
const long long mod = 1e9 + 7;
long long ster[1010][1010];
long long psum[1010];
int m, n;
bool same[1010][1010];
char str[1010];
bool check[1010];
int dfs(int now) {
  int ans = 1;
  for (int i = 0; i < m; i++)
    if (!check[i] && same[now][i]) {
      check[i] = true;
      ans += dfs(i);
    }
  return ans;
}
int main() {
  scanf("%d %d", &m, &n);
  ster[1][1] = 1, psum[1] = 1;
  for (int i = 2; i <= m; i++)
    for (int j = 1; j <= m; j++) {
      ster[i][j] = (j * ster[i - 1][j] % mod + ster[i - 1][j - 1]) % mod;
      psum[i] = (psum[i] + ster[i][j]) % mod;
    }
  for (int i = 0; i < m; i++)
    for (int j = 0; j < m; j++) same[i][j] = true;
  for (int i = 0; i < n; i++) {
    scanf("%s", str);
    for (int j = 0; j < m; j++)
      for (int k = 0; k < m; k++)
        if (str[j] != str[k]) same[j][k] = same[k][j] = false;
  }
  long long ans = 1;
  for (int i = 0; i < m; i++) {
    if (!check[i]) {
      check[i] = true;
      ans = ans * psum[dfs(i)] % mod;
    }
  }
  printf("%I64d", ans);
  return 0;
}
