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;
int r[1050];
int root(int x) { return (x == r[x]) ? x : (r[x] = root(r[x])); }
char in[55][1050];
int sz[1050];
long long dp[1050];
long long C[1050][1050];
int main() {
  int M, N, i, j, k;
  scanf("%d %d", &M, &N);
  for (i = 0; i <= 1000; i++) {
    C[i][0] = C[i][i] = 1;
    for (j = 1; j < i; j++)
      C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % 1000000007;
  }
  dp[0] = dp[1] = 1;
  for (i = 2; i <= 1000; i++) {
    dp[i] = 0;
    for (j = 0; j <= i - 1; j++)
      dp[i] = (dp[i] + dp[i - 1 - j] * C[i - 1][j]) % 1000000007;
  }
  for (i = 0; i < N; i++) scanf("%s", in[i]);
  for (i = 0; i < M; i++) r[i] = i;
  for (i = 0; i < M; i++) {
    for (j = 0; j < i; j++) {
      for (k = 0; k < N; k++)
        if (in[k][i] != in[k][j]) break;
      if (k >= N) r[root(i)] = root(j);
    }
  }
  int c = 0;
  for (i = 0; i < M; i++) sz[root(i)]++;
  long long ans = 1;
  for (i = 0; i < M; i++) ans = ans * dp[sz[i]] % 1000000007;
  return !printf("%lld\n", ans);
}
