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 MOD = 1e9 + 7;
int m, n;
char s[1009][1009];
long long ans = 1;
bool vis[1009];
long long B[1009], C[1009][1009];
string doit(int x, int y) {
  string ans = "";
  if (s[x][y] != '1')
    return ans;
  else {
    for (int j = 0; j < m; j++) ans.push_back(s[x][j]);
    return ans;
  }
  return ans;
}
string solve(int x, int y) {
  string ans = doit(x, y);
  if ((int)ans.size() != 0) return ans;
  for (int j = 0; j < m; j++) {
    if (s[x][j] == '0')
      s[x][j] = '1';
    else
      s[x][j] = '0';
  }
  return doit(x, y);
}
string update(string x, string y) {
  string ans = "";
  for (int j = 0; j < m; j++) {
    if (x[j] == '1' && y[j] == '1')
      ans.push_back('1');
    else
      ans.push_back('0');
  }
  return ans;
}
void getans(string &s) {
  int cnt = 0;
  for (int j = 0; j < m; j++) {
    if (s[j] == '1') {
      cnt++;
      vis[j] = 1;
    }
  }
  ans = ans * B[cnt] % MOD;
}
int main() {
  for (int i = 0; i < 1009; i++) {
    C[i][i] = C[i][0] = 1;
    for (int j = 1; j < i; j++) C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % MOD;
  }
  B[0] = 1;
  for (int i = 1; i < 1009; i++) {
    for (int j = 1; j <= i; j++) {
      B[i] = (B[i] + C[i - 1][j - 1] * B[i - j]) % MOD;
    }
  }
  cin >> m >> n;
  for (int i = 1; i <= n; i++) {
    scanf("%s", s[i]);
  }
  for (int i = 0; i < m; i++) {
    if (vis[i]) continue;
    bool ok = 1;
    string tmp = "";
    for (int j = 0; j < m; j++) tmp.push_back('1');
    for (int j = 1; j <= n; j++) {
      string cur = solve(j, i);
      int sz = cur.size();
      if (sz == 0) {
        ok = 0;
        break;
      }
      tmp = update(tmp, cur);
    }
    if (ok) {
      getans(tmp);
    }
  }
  cout << ans << endl;
  return 0;
}
