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;
bool gg[2000];
int sz[2000], par[2000];
char s[100][2000];
int dp[2000][2000];
int C(int n, int m) {
  if (n < m || m < 0) return 0;
  if (n == 0 || n == m) return 1;
  if (dp[n][m]) return dp[n][m];
  if (n - m < m) return dp[n][m] = C(n, n - m);
  return dp[n][m] = (C(n - 1, m - 1) + C(n - 1, m)) % 1000000007;
}
int main() {
  int m, n, r = 1;
  scanf("%d%d", &m, &n);
  for (int i = 0; i < n; i++) scanf("%s", s[i]);
  par[0] = 1;
  for (int i = 1; i <= m; i++) {
    for (int j = 0; j < i; j++) {
      par[i] = (par[i] + 1LL * par[j] * C(i - 1, j) % 1000000007) % 1000000007;
    }
  }
  for (int i = 0; i < m; i++) {
    for (int j = 0; j < i; j++) {
      if (gg[j]) continue;
      gg[i] = true;
      for (int k = 0; k < n; k++) {
        if (s[k][j] != s[k][i]) {
          gg[i] = false;
          break;
        }
      }
      if (gg[i]) {
        sz[j]++;
        break;
      }
    }
    if (!gg[i]) sz[i] = 1;
  }
  for (int i = 0; i < m; i++) {
    if (!gg[i]) r = 1LL * r * par[sz[i]] % 1000000007;
  }
  printf("%d\n", r);
  return 0;
}
