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 long long MOD = 1000000007;
string andstring(const string& a, const string& b) {
  string res(a);
  for (int i = 0; i < b.size(); i++)
    res[i] = ((a[i] - '0') & (b[i] - '0')) + '0';
  return res;
}
string invstring(const string& a) {
  string res(a);
  for (int i = 0; i < a.size(); i++) res[i] = '0' + '1' - res[i];
  return res;
}
const int MAXM = 1001;
long long dp[MAXM];
long long cnk[MAXM][MAXM];
int main() {
  int m, n;
  cin >> m >> n;
  vector<string> s(n);
  for (int i = 0; i < n; i++) cin >> s[i];
  for (int i = 0; i < n; i++) s.push_back(invstring(s[i]));
  sort(s.begin(), s.end());
  s.erase(unique(s.begin(), s.end()), s.end());
  n = s.size();
  vector<string> minstring(m);
  for (int i = 0; i < m; i++) {
    minstring[i] = string(m, '1');
    for (int j = 0; j < n; j++)
      if (s[j][i] == '1') minstring[i] = andstring(minstring[i], s[j]);
  }
  sort(minstring.begin(), minstring.end());
  minstring.erase(unique(minstring.begin(), minstring.end()), minstring.end());
  for (int i = 0; i <= m; i++) {
    cnk[i][0] = 1;
    for (int j = 1; j < i; j++)
      cnk[i][j] = (cnk[i - 1][j - 1] + cnk[i - 1][j]) % MOD;
    cnk[i][i] = 1;
  }
  dp[0] = 1;
  for (int i = 1; i <= m; i++) {
    dp[i] = 0;
    for (int j = 0; j < i; j++)
      dp[i] = (dp[i] + dp[i - j - 1] * cnk[i - 1][j] % MOD) % MOD;
  }
  long long res = 1;
  for (int i = 0; i < minstring.size(); i++) {
    int sum = 0;
    for (int j = 0; j < m; j++) sum += minstring[i][j] == '1' ? 1 : 0;
    res = res * dp[sum] % MOD;
  }
  printf("%lld\n", res);
  return 0;
}
