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 Maxn = 1E3 + 5;
const int Mod = 1E9 + 7;
int N, M;
char a[55][Maxn];
int T[Maxn];
int siz[Maxn];
int C[Maxn][Maxn];
int f[Maxn];
int Ans;
int Find(int X) { return (~T[X] ? T[X] = Find(T[X]) : X); }
void Union(int X, int Y) {
  int Fx = Find(X), Fy = Find(Y);
  if (Fx != Fy) T[Fx] = Fy, siz[Fy] += siz[Fx];
}
void PrePare() {
  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 - 1] + C[i - 1][j]) % Mod;
  }
  f[0] = 1;
  for (int i = 1; i <= N; ++i)
    for (int j = 1; j <= i; ++j)
      f[i] = (f[i] + (long long)C[i - 1][j - 1] * f[i - j]) % Mod;
}
int main() {
  scanf("%d %d", &N, &M);
  for (int i = 1; i <= M; ++i) scanf("%s", a[i] + 1);
  PrePare();
  memset(T, -1, sizeof(T));
  for (int i = 1; i <= N; ++i) siz[i] = 1;
  for (int i = 1; i <= N; ++i)
    for (int j = i + 1, k; j <= N; ++j) {
      for (k = 1; k <= M; ++k)
        if (a[k][i] != a[k][j]) break;
      if (k > M) Union(i, j);
    }
  Ans = 1;
  for (int i = 1; i <= N; ++i)
    if (T[i] == -1) Ans = (long long)Ans * f[siz[i]] % Mod;
  printf("%d", Ans);
  return 0;
}
