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;
int n, m;
char t[52][1002];
int dp[1002][1002], Bell[1002];
long long v[1002];
int ans;
void compDp() {
  dp[0][0] = 1;
  for (int i = 1; i <= m; ++i) {
    for (int d = 1; d <= i; ++d) {
      dp[i][d] = (1LL * dp[i - 1][d] * d + dp[i - 1][d - 1]) % 1000000007;
      Bell[i] += dp[i][d];
      if (Bell[i] >= 1000000007) Bell[i] -= 1000000007;
    }
  }
}
int main() {
  scanf("%d%d", &m, &n);
  for (int i = 1; i <= n; ++i) {
    bool add = 1;
    scanf("\n%s", t[i]);
    for (int j = 1; j < i; ++j) {
      bool comp = 1;
      for (int b = 0; comp && b < m; ++b)
        if (t[i][b] == t[j][b]) comp = 0;
      if (comp) {
        add = 0;
        break;
      }
    }
    if (add)
      for (int b = 0; b < m; ++b) v[b] = 2LL * v[b] + t[i][b] - '0';
  }
  compDp();
  sort(v, v + m);
  int ans = 1, cnt = 1;
  v[m] = -1;
  for (int i = 1; i <= m; ++i)
    if (v[i] != v[i - 1]) {
      ans = (1LL * ans * Bell[cnt]) % 1000000007;
      cnt = 1;
    } else
      ++cnt;
  printf("%d\n", ans);
  return 0;
}
