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 bool cmin(T& a, const T& b) {
  return a > b ? a = b, 1 : 0;
}
template <typename T>
inline bool cmax(T& a, const T& b) {
  return a < b ? a = b, 1 : 0;
}
int read() {
  int x = 0, f = 1;
  char ch;
  for (ch = getchar(); !isdigit(ch); ch = getchar())
    if (ch == '-') f = -1;
  for (; isdigit(ch); x = x * 10 + ch - '0', ch = getchar())
    ;
  return x * f;
}
const int MaxN = 1234, P = 1e9 + 7;
int m, n;
long long b[MaxN], c[MaxN][MaxN], f[MaxN];
map<long long, long long> g;
int main() {
  int i, j;
  m = read();
  n = read();
  while (n--)
    for (j = 1; j <= m; j++) {
      long long x;
      scanf("%1lld", &x);
      b[j] |= (x << n);
    }
  for (i = 1; i <= m; i++) ++g[b[i]];
  c[0][0] = 1;
  for (i = 1; i <= m; i++) {
    c[i][0] = 1;
    for (j = 1; j <= i; j++) c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % P;
  }
  f[0] = f[1] = 1;
  for (i = 2; i <= m; i++)
    for (j = 0; j < i; j++) f[i] = (f[i] + (long long)f[j] * c[i - 1][j]) % P;
  int ans = 1;
  for (map<long long, long long>::iterator it = g.begin(); it != g.end(); it++)
    ans = (long long)ans * f[it->second] % P;
  cout << ans << endl;
  return 0;
}
