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 = 1e9 + 7;
void imult(long long& lhs, const long long& rhs) {
  lhs = ((lhs % mod) * (rhs % mod)) % mod;
}
void iadd(long long& lhs, const long long& rhs) {
  lhs = ((lhs % mod) + (rhs % mod)) % mod;
}
void isubt(long long& lhs, const long long& rhs) {
  long long ret = (lhs - rhs) % mod;
  lhs = ret + (ret < 0 ? mod : 0);
}
long long bpw(const long long& lhs, const long long& rhs) {
  if (rhs == 0) return 1;
  long long ret = bpw(lhs, rhs >> 1);
  imult(ret, ret);
  if (rhs & 1) imult(ret, lhs);
  return ret;
}
long long inv(const long long& x) { return bpw(x, mod - 2); }
long long mult(const long long& lhs, const long long& rhs) {
  return ((lhs % mod) * (rhs % mod)) % mod;
}
long long add(const long long& lhs, const long long& rhs) {
  return ((lhs % mod) + (rhs % mod)) % mod;
}
long long subt(const long long& lhs, const long long& rhs) {
  long long ret = (lhs - rhs) % mod;
  return ret + (ret < 0 ? mod : 0);
}
long long div(const long long& lhs, const long long& rhs) {
  return mult(lhs, inv(rhs));
}
const int sz = 1e3;
int m;
bool a[sz][sz];
long long part[sz + 5];
int chs[sz + 1][sz + 1];
vector<int> adj[sz];
bitset<sz> vis;
int dfs(int src) {
  vis[src] = 1;
  int tot = 1;
  for (int x : adj[src]) {
    if (!vis[x]) tot += dfs(x);
  }
  return tot;
}
void calcBell() {
  for (int i = 0; i <= sz; i++) {
    for (int j = 0; j <= i; j++) {
      if (!j)
        chs[i][j] = 1;
      else
        chs[i][j] = add(chs[i - 1][j], chs[i - 1][j - 1]);
    }
  }
  part[0] = 1;
  for (int i = 1; i <= sz; i++) {
    for (int j = 0; j < i; j++) iadd(part[i], mult(chs[i - 1][j], part[j]));
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  string s;
  cin >> m >> n;
  while (n--) {
    cin >> s;
    for (int i = 0; i < m; i++) {
      for (int j = i + 1; j < m; j++) {
        if (s[i] != s[j]) {
          a[i][j] = a[j][i] = 1;
        }
      }
    }
  }
  for (int i = 0; i < m; i++) {
    for (int j = i + 1; j < m; j++) {
      if (!a[i][j]) {
        adj[i].push_back(j);
        adj[j].push_back(i);
      }
    }
  }
  long long ret = 1;
  calcBell();
  for (int i = 0; i < m; i++) {
    int tmp = dfs(i);
    imult(ret, part[tmp]);
  }
  cout << ret << '\n';
  return 0;
}
