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;
template <typename T>
inline void Read(T &x) {
  char c = getchar();
  bool f = false;
  for (x = 0; !isdigit(c); c = getchar()) {
    if (c == '-') {
      f = true;
    }
  }
  for (; isdigit(c); c = getchar()) {
    x = x * 10 + c - '0';
  }
  if (f) {
    x = -x;
  }
}
template <typename T>
inline bool CheckMax(T &a, const T &b) {
  return a < b ? a = b, true : false;
}
template <typename T>
inline bool CheckMin(T &a, const T &b) {
  return a > b ? a = b, true : false;
}
const int N = 1005;
const int M = 55;
const int mod = 1e9 + 7;
int n, m, f[N], g[N], h[N], c[N][N];
char s[M][N];
inline int F(int x) {
  while (x != f[x]) {
    x = f[x] = f[f[x]];
  }
  return x;
}
int main() {
  Read(n), Read(m);
  for (int i = 0; i < m; ++i) {
    scanf("%s", s[i]);
  }
  for (int i = 0; i <= n; ++i) {
    c[i][0] = 1;
    for (int j = 1; j <= i; ++j) {
      c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % mod;
    }
  }
  g[0] = g[1] = 1;
  for (int i = 2; i <= n; ++i) {
    for (int j = 0; j < i; ++j) {
      g[i] = (1LL * g[i - j - 1] * c[i - 1][j] + g[i]) % mod;
    }
  }
  for (int i = 0; i < n; ++i) {
    f[i] = i;
  }
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < i; ++j) {
      if (F(i) != F(j)) {
        bool flg = true;
        for (int k = 0; k < m; ++k) {
          if (s[k][i] != s[k][j]) {
            flg = false;
            break;
          }
        }
        if (flg) {
          f[F(i)] = F(j);
        }
      }
    }
  }
  for (int i = 0; i < n; ++i) {
    ++h[F(i)];
  }
  int ans = 1;
  for (int i = 0; i < n; ++i) {
    if (F(i) == i) {
      ans = 1LL * ans * g[h[i]] % mod;
    }
  }
  printf("%d\n", ans);
  return 0;
}
