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 upmin(T &x, T y) {
  y < x ? x = y : 0;
}
template <typename T>
inline void upmax(T &x, T y) {
  x < y ? x = y : 0;
}
const long double pi = acos(-1);
const int oo = 1 << 30;
const long long OO = 1e18;
const int N = 1010;
const int mod = 1e9 + 7;
int gi() {
  int w = 0;
  bool q = 1;
  char c = getchar();
  while ((c < '0' || c > '9') && c != '-') c = getchar();
  if (c == '-') q = 0, c = getchar();
  while (c >= '0' && c <= '9') w = w * 10 + c - '0', c = getchar();
  return q ? w : -w;
}
int S[N][N], C[N][N];
char s[N][N];
int siz[N];
int main() {
  int n = gi(), m = gi(), i, j, k, ans = 1, sum;
  S[0][0] = 1;
  for (i = 0; i <= n; i++)
    for (j = C[i][0] = 1; j <= i; j++)
      C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % mod;
  for (i = 1; i <= n; i++)
    for (j = 1; j <= i; j++)
      S[i][j] = (S[i - 1][j - 1] + 1LL * S[i - 1][j] * j) % mod;
  for (i = 0; i < m; i++) scanf("%s", s[i]);
  for (i = 0; i < n; i++) {
    for (j = 0; j < i; j++) {
      for (k = 0; k < m; k++)
        if (s[k][i] != s[k][j]) break;
      if (k == m) {
        ++siz[j];
        break;
      }
    }
    if (j == i) ++siz[i];
  }
  for (i = 0; i < n; i++)
    if (siz[i]) {
      sum = 0;
      for (j = 1; j <= siz[i]; j++) (sum += S[siz[i]][j]) %= mod;
      ans = 1LL * ans * sum % mod;
    }
  cout << ans << endl;
  return 0;
}
