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;
using ll = long long;
using ull = unsigned long long;
const ll mod = 1000000007;
int m;
vector<vector<bool>> g;
vector<bool> used;
int dfs(int v) {
  int sum = 1;
  used[v] = true;
  for (int u = 0; u < m; ++u) {
    if (!used[u] && g[v][u]) sum += dfs(u);
  }
  return sum;
}
void solve() {
  int n;
  scanf("%d%d\n", &m, &n);
  vector<vector<ll>> c(m + 1, vector<ll>(m + 1, 0));
  c[0][0] = 1;
  for (int i = 1; i <= m; ++i) {
    c[i][0] = 1;
    for (int j = 1; j <= m; ++j)
      c[i][j] = (c[i][j] + c[i - 1][j] + c[i - 1][j - 1]) % mod;
  }
  vector<vector<ll>> dp(m + 1, vector<ll>(m + 1, 0));
  dp[0][0] = 1;
  for (int i = 1; i <= m; ++i) {
    for (int j = 1; j <= i; ++j) {
      dp[i][j] = (dp[i - 1][j] * j + dp[i - 1][j - 1]) % mod;
    }
  }
  g.assign(m, vector<bool>(m, true));
  for (int i = 0; i < n; ++i) {
    string s;
    getline(cin, s);
    for (int j = 0; j < m; ++j) {
      for (int k = 0; k < j; ++k) {
        if (s[j] != s[k]) g[k][j] = g[j][k] = false;
      }
    }
  }
  used.assign(m, false);
  vector<ll> cs;
  for (int i = 0; i < m; ++i) {
    if (!used[i]) cs.push_back(dfs(i));
  }
  ll answer = 1;
  for (int i = 0; i < cs.size(); ++i) {
    ll sum = 0;
    for (int j = 1; j <= cs[i]; ++j) sum += dp[cs[i]][j];
    answer = (answer * (sum % mod)) % mod;
  }
  printf("%lld", answer);
}
int main() {
  solve();
  return 0;
}
