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;
using ll = long long;
int mod = 1000000000 + 7;
const int maxm = 1000 + 100;
int comb[maxm][maxm];
int bell[maxm];
ll digit[1000];
void init_comb() {
  comb[0][0] = 1;
  for (int i = 1; i < maxm; i++) {
    comb[i][0] = 1;
    for (int j = 1; j <= i; j++) {
      comb[i][j] = (comb[i - 1][j - 1] + comb[i - 1][j]) % mod;
    }
  }
}
void init_bell() {
  bell[0] = 1;
  for (int i = 0; i + 1 < maxm; i++) {
    bell[i + 1] = 0;
    for (int j = 0; j <= i; j++) {
      bell[i + 1] = (bell[i + 1] + 1ll * bell[j] * comb[i][j]) % mod;
    }
  }
}
int main() {
  init_comb();
  init_bell();
  int m, n;
  cin >> m >> n;
  string str;
  for (int j = 0; j < m; j++) digit[j] = 0;
  for (int i = 0; i < n; i++) {
    cin >> str;
    for (int j = 0; j < m; j++) {
      if (str[j] == '1') digit[j] |= 1ll << i;
    }
  }
  sort(digit, digit + m);
  int blocks = 1;
  ll ans = 1;
  for (int i = 1; i < m; i++) {
    if (digit[i] == digit[i - 1]) {
      blocks++;
    } else {
      ans = ans * bell[blocks] % mod;
      blocks = 1;
    }
  }
  ans = ans * bell[blocks] % mod;
  printf("%lld\n", ans);
}
