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 m, n;
char s[51][1010];
int id[1010], cnt[1010], c = 0;
long long invf[1010];
long long powm(long long x, int e) {
  long long prod = 1;
  while (e) {
    if (e & 1) prod = (prod * x) % 1000000007;
    x = (x * x) % 1000000007;
    e >>= 1;
  }
  return prod;
}
long long inv(long long x) { return powm(x, 1000000007 - 2); }
long long calc(int e) {
  long long p[1010], sum = 0;
  int i, j;
  for (i = 1; i <= e; ++i) {
    p[i] = (powm(i, e) * invf[i]) % 1000000007;
    for (j = 1; j < i; ++j) {
      p[i] =
          (p[i] + 1000000007 - (p[j] * invf[i - j]) % 1000000007) % 1000000007;
    }
    sum += p[i];
  }
  return sum % 1000000007;
}
int main() {
  int i, j, k;
  long long tot = 1;
  scanf("%d%d", &m, &n);
  for (i = 1; i <= n; ++i) {
    scanf("%s", s[i]);
  }
  invf[1] = 1;
  for (i = 2; i <= m; ++i) {
    invf[i] = (invf[i - 1] * inv(i)) % 1000000007;
  }
  for (j = 0; j < m; ++j) {
    if (!id[j]) {
      id[j] = ++c;
      cnt[c] = 1;
    }
    for (k = j + 1; k < m; ++k) {
      if (id[k]) continue;
      for (i = 1; i <= n; ++i) {
        if (s[i][j] != s[i][k]) break;
      }
      if (i == n + 1) {
        id[k] = id[j];
        cnt[id[j]]++;
      }
    }
  }
  for (i = 1; i <= c; ++i) {
    tot = (tot * calc(cnt[i])) % 1000000007;
  }
  printf("%I64d", tot);
  return 0;
}
