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 = (int)1e9;
const long long linf = (long long)1e18;
const int mod = (int)1e9 + 7;
const long double eps = (long double)1e-8;
const int maxn = (int)1e5 + 5;
const long double pi = acos(-1);
int m, n;
vector<int> a[55];
int c[1111], k[1111], ccnt;
long long ans;
long long dres[1111], d[1111][1111];
void dcalc() {
  d[0][0] = 1;
  for (auto i = 1; i < m + 1; ++i) {
    for (auto j = 1; j < m + 1; ++j) {
      d[i][j] = (d[i - 1][j] * j + d[i - 1][j - 1]) % mod;
    }
  }
  memset((dres), 0, sizeof(dres));
  for (auto i = 1; i < m + 1; ++i) {
    for (auto j = 0; j < i + 1; ++j) dres[i] = (dres[i] + d[i][j]) % mod;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> m >> n;
  for (auto i = 0; i < n; ++i) {
    string s;
    cin >> s;
    for (auto j = 0; j < m; ++j) {
      a[i].push_back((int)(s[j] - '0'));
    }
  }
  memset((c), 0, sizeof(c));
  ccnt = 0;
  for (auto i = 0; i < m; ++i) {
    if (c[i]) continue;
    ++ccnt;
    c[i] = ccnt;
    k[ccnt] = 1;
    for (auto j = i + 1; j < m; ++j) {
      if (c[j]) continue;
      int fl = 0;
      for (auto k = 0; k < n; ++k) {
        if (a[k][i] != a[k][j]) {
          fl = 1;
          break;
        }
      }
      if (!fl) {
        ++k[ccnt];
        c[j] = ccnt;
      }
    }
  }
  dcalc();
  ans = 1;
  for (auto i = 1; i < ccnt + 1; ++i) {
    ans = (ans * dres[k[i]]) % mod;
  }
  cout << ans << endl;
  return 0;
}
