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 n, m, t[2000];
long long ans, dp[2000][2000], DP[2000];
string s[100];
bool ok(int i, int j) {
  for (int x = 0; x < n; x++)
    if (s[x][i] != s[x][j]) return false;
  return true;
}
int main() {
  cin >> m >> n;
  for (int i = 0; i < n; i++) cin >> s[i];
  for (int j = 0; j < m; j++) {
    t[j]++;
    for (int i = 0; i < j; i++)
      if (ok(i, j)) {
        t[i]++;
        t[j]--;
        break;
      }
  }
  DP[0] = dp[0][0] = 1;
  for (int i = 1; i <= 1000; i++)
    for (int j = 1; j <= i; j++) {
      dp[i][j] = (dp[i - 1][j - 1] + dp[i - 1][j] * j) % MOD;
      DP[i] = (DP[i] + dp[i][j]) % MOD;
    }
  ans = 1;
  for (int i = 0; i < m; i++) ans = (ans * DP[t[i]]) % MOD;
  cout << ans << endl;
}
