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 N = 200031;
bool is_good(int mask, int sz) {
  set<int> S;
  vector<int> V;
  for (int i = 0; i < (1 << sz); i++) {
    if (mask & (1 << i)) {
      S.insert(i);
      V.push_back(i);
    }
  }
  for (int i = 0; i < V.size(); i++) {
    if (S.find(V[i] ^ ((1 << sz) - 1)) == S.end()) return false;
  }
  for (int i = 0; i < V.size(); i++) {
    for (int j = i + 1; j < V.size(); j++) {
      if (S.find(V[i] & V[j]) == S.end()) return false;
    }
  }
  return true;
}
void show_set(int mask, int n) {
  cout << "---" << endl;
  for (int i = 0; i < (1 << n); i++) {
    if (mask & (1 << i)) {
      for (int j = n - 1; j >= 0; j--) {
        if (i & (1 << j))
          cout << 1;
        else
          cout << 0;
      }
      cout << endl;
    }
  }
}
int len, n;
string st[N];
int used[N];
int C;
void dfs(int v) {
  used[v] = 1;
  ++C;
  for (int i = 0; i < len; i++) {
    if (used[i]) continue;
    int can_move = 1;
    for (int j = 0; j < n; j++) {
      if (st[j][v] != st[j][i]) can_move = 0;
    }
    if (can_move) dfs(i);
  }
}
long long dp[1031][1031], BELL[1031];
int main() {
  ios_base::sync_with_stdio(0);
  dp[1][1] = 1;
  for (int i = 1; i <= 1000; i++) {
    for (int j = 0; j <= 1000; j++) {
      if (dp[i][j] == 0) continue;
      dp[i + 1][j + 1] += dp[i][j];
      dp[i + 1][j] += dp[i][j] * j;
      dp[i + 1][j + 1] %= 1000000007;
      dp[i + 1][j] %= 1000000007;
    }
  }
  for (int i = 1; i <= 1000; i++) {
    for (int j = 1; j <= 1000; j++) {
      BELL[i] += dp[i][j];
      BELL[i] %= 1000000007;
    }
  }
  cin >> len >> n;
  for (int i = 0; i < n; i++) {
    cin >> st[i];
  }
  long long ans = 1;
  for (int i = 0; i < len; i++) {
    if (used[i]) continue;
    C = 0;
    dfs(i);
    ans *= BELL[C];
    ans %= 1000000007;
  }
  cout << ans << endl;
  cin.get();
  cin.get();
  return 0;
}
