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 n, m;
bitset<2010> g[2010];
long long f[2010][2010];
map<long long, int> cnt;
long long calc(int n) {
  long long ans = 0;
  for (int i = 0; i <= n; i++) ans = (ans + f[n][i]) % 1000000007LL;
  return ans;
}
long long solve(int n, int m) {
  long long ans = 1LL;
  int k = 1;
  for (int i = 0; i <= m; i++) f[i][i] = 1;
  for (int i = 1; i <= m; i++) {
    for (int j = 1; j < i; j++)
      f[i][j] =
          (j * f[i - 1][j] % 1000000007LL + f[i - 1][j - 1]) % 1000000007LL;
  }
  for (int i = 1; i <= m; i++) {
    int t = 0;
    for (int j = k; j <= n; j++)
      if (g[j][i]) {
        t = j;
        break;
      }
    if (!t) continue;
    swap(g[t], g[k]);
    for (int j = 1; j <= n; j++)
      if (j != k && g[j][i]) g[j] ^= g[k];
    k++;
  }
  for (int i = 1; i <= m; i++) {
    long long now = 0;
    for (int j = 1; j <= n; j++)
      if (g[j][i]) now |= (1LL << (j - 1));
    cnt[now]++;
  }
  for (map<long long, int>::iterator ii = cnt.begin(); ii != cnt.end(); ii++)
    if ((*ii).first) ans = ans * calc((*ii).second) % 1000000007LL;
  return ans;
}
int main() {
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    for (int j = 1; j <= n; j++) {
      char ch;
      while (!isdigit(ch = getchar()))
        ;
      g[i][j] = ch - '0';
    }
  }
  m++;
  for (int j = 1; j <= n; j++) g[m][j] = 1;
  for (int i = m + 1; i <= 2 * m; i++) {
    for (int j = 1; j <= n; j++) g[i][j] = g[i - m][j] ^ 1;
  }
  cout << solve(2 * m, n) << endl;
  return 0;
}
