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;
const int mod = 1000000007;
const int N = 1100;
const int M = 60;
char s[M][N];
int n, m, rev[N], vis[N];
long long c[N][N], f[N];
int main() {
  int i, j, k, tot;
  scanf("%d%d", &n, &m);
  c[0][0] = 1;
  for (i = 1; i <= n; i++) c[i][0] = 1;
  for (i = 1; i <= n; i++)
    for (j = 1; j <= i; j++) c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % mod;
  f[0] = f[1] = 1;
  for (i = 2; i <= n; i++)
    for (j = 0; j < i; j++) f[i] = (f[i] + f[j] * c[i - 1][j]) % mod;
  for (i = 1; i <= m; i++) scanf("%s", s[i] + 1);
  long long ans = 1;
  for (i = 1; i <= n; i++) {
    if (vis[i]) continue;
    for (j = 1; j <= m; j++)
      if ((s[j][i] - '0') ^ rev[j] == 0) rev[j] ^= 1;
    tot = 0;
    for (j = i; j <= n; j++)
      if (!vis[j]) {
        bool fl = 1;
        for (k = 1; k <= m; k++) fl &= (s[k][j] - '0') ^ rev[k];
        if (fl) {
          tot++;
          vis[j] = 1;
        }
      }
    ans = (ans * f[tot]) % mod;
  }
  printf("%d\n", ans);
  return 0;
}
