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 = 50;
const int M = 1001;
const int MOD = (int)1e9 + 7;
using Bitset = std::bitset<M>;
int n, m, s[M][M], bell[M];
char buf[M];
bool vis[M];
Bitset st[N], st2[N];
void update(int& x, int a) {
  x += a;
  if (x >= MOD) {
    x -= MOD;
  }
}
int main() {
  s[0][0] = 1;
  for (int i = 1; i < M; ++i) {
    bell[i] = 0;
    for (int j = 1; j <= i; ++j) {
      s[i][j] = (1LL * s[i - 1][j] * j + s[i - 1][j - 1]) % MOD;
      update(bell[i], s[i][j]);
    }
  }
  scanf("%d%d", &m, &n);
  for (int i = 0; i < n; ++i) {
    scanf("%s", buf);
    for (int j = 0; j < m; ++j) {
      if (buf[j] == '1') {
        st[i].set(j);
      }
    }
    st2[i] = st[i];
    st2[i].flip();
  }
  int result = 1;
  for (int i = 0; i < m; ++i) {
    if (!vis[i]) {
      Bitset b;
      b.flip();
      for (int j = 0; j < n; ++j) {
        if (st[j].test(i)) {
          b &= st[j];
        } else {
          b &= st2[j];
        }
      }
      int sz = 0;
      for (int j = 0; j < m; ++j) {
        if (b.test(j)) {
          vis[j] = true;
          sz++;
        }
      }
      result = 1LL * result * bell[sz] % MOD;
    }
  }
  printf("%d\n", result);
}
