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 INF = 1e9 + 7;
long long dp[1005][1005];
string S[55];
vector<int> G[1005];
int sz;
bool vis[1005];
void dfs(int v) {
  vis[v] = true;
  sz++;
  for (int k = 0; k < (G[v].size()); k++) {
    int u = G[v][k];
    if (!vis[u]) dfs(u);
  }
}
int main() {
  ios::sync_with_stdio(false);
  int n, m;
  cin >> n >> m;
  dp[0][0] = 1;
  for (int i = 1; i <= (n); i++)
    for (int j = 1; j <= (i); j++)
      dp[i][j] = (dp[i - 1][j] * j + dp[i - 1][j - 1]) % INF;
  for (int k = 0; k < (m); k++) cin >> S[k];
  for (int i = 0; i < (n); i++)
    for (int j = 0; j < (n); j++) {
      bool ok = true;
      for (int k = 0; k < (m); k++)
        if (S[k][i] != S[k][j]) ok = false;
      if (ok) G[i].push_back(j);
    }
  long long ans = 1LL;
  for (int k = 0; k < (n); k++)
    if (!vis[k]) {
      sz = 0;
      dfs(k);
      long long ca = 0LL;
      for (int k = 1; k <= (sz); k++) ca += dp[sz][k];
      ca %= INF;
      ans = ans * ca % INF;
    }
  cout << ans << '\n';
  return 0;
}
