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>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
template <class T, class U>
inline void Max(T &a, U b) {
  if (a < b) a = b;
}
template <class T, class U>
inline void Min(T &a, U b) {
  if (a > b) a = b;
}
inline void add(int &a, int b) {
  a += b;
  while (a >= 1000000007) a -= 1000000007;
}
int pow(int a, int b) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = ans * (long long)a % 1000000007;
    a = (long long)a * a % 1000000007;
    b >>= 1;
  }
  return ans;
}
int f[1005], dp[1005][1005];
long long v[1005];
char s[1005];
int main() {
  int j, i, k, T, ca = 0, K = 0, m = 0, n;
  scanf("%d%d", &m, &n);
  for (int i = 0; i < n; i++) {
    scanf("%s", s);
    for (int k = 0; k < m; k++)
      if (s[k] == '1') {
        v[k] |= 1LL << i;
      }
  }
  sort(v, v + m);
  dp[0][0] = 1;
  for (int i = 0; i < m; i++) {
    for (int j = 0; j < i + 1; j++) {
      add(dp[i + 1][j], 1LL * dp[i][j] * j % 1000000007);
      add(dp[i + 1][j + 1], dp[i][j]);
    }
  }
  for (int i = 1; i < m + 1; i++)
    for (int j = 1; j < i + 1; j++) add(f[i], dp[i][j]);
  int ans = 1;
  for (int i = 0; i < m; i++) {
    j = i;
    while (j < m && v[j] == v[i]) j++;
    ans = 1LL * ans * f[j - i] % 1000000007;
    i = j - 1;
  }
  printf("%d\n", ans);
  return 0;
}
