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 int N = 1010;
const int M = 60;
const int moder = 1e9 + 7;
std::bitset<M> set[N];
int dp[N];
char s[N][N];
int cnt[N];
int comb[N][N];
int main() {
  for (int i = 0; i < N; ++i) {
    comb[i][0] = 1;
    for (int j = 1; j <= i; ++j) {
      comb[i][j] = (comb[i - 1][j - 1] + comb[i - 1][j]) % moder;
    }
  }
  dp[0] = 1;
  for (int i = 1; i < N; ++i) {
    for (int j = 1; j <= i; ++j) {
      dp[i] = (dp[i] + 1ll * dp[i - j] * comb[i - 1][j - 1]) % moder;
    }
  }
  int m, n;
  scanf("%d%d", &m, &n);
  for (int i = 0; i < n; ++i) {
    scanf("%s", s[i]);
  }
  for (int i = 0; i < m; ++i) {
    for (int j = 0; j < n; ++j) {
      if (s[j][i] == '1') {
        set[i].set(j);
      }
    }
  }
  for (int i = 0; i < m; ++i) {
    for (int j = 0; j <= i; ++j) {
      if (set[i] == set[j]) {
        ++cnt[j];
        break;
      }
    }
  }
  int ans = 1;
  for (int i = 0; i < m; ++i) {
    ans = 1ll * ans * dp[cnt[i]] % moder;
  }
  return printf("%d\n", ans), 0;
}
