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;
template <class T>
int getbit(T s, int i) {
  return (s >> i) & 1;
}
template <class T>
T onbit(T s, int i) {
  return s | (T(1) << i);
}
template <class T>
T offbit(T s, int i) {
  return s & (~(T(1) << i));
}
template <class T>
int cntbit(T s) {
  return __builtin_popcount(s);
}
template <class T>
T gcd(T a, T b) {
  T r;
  while (b != 0) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
template <class T>
T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
int n, m;
long long a[2005][2005];
long long f[2005], C[2005][2005], flag[2005];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  memset(C, 0, sizeof(C));
  for (int i = 0; i < (2005); ++i) C[0][i] = 1;
  for (int i = (1); i <= (2005 - 1); ++i)
    for (int j = 1; j <= i; j++)
      C[j][i] = (C[j - 1][i - 1] + C[j][i - 1]) % 1000000007;
  f[1] = 1;
  f[0] = 1;
  for (int i = 2; i <= 2005 - 1; i++) {
    f[i] = 0;
    for (int j = 1; j <= i; j++) {
      f[i] = (f[i] + C[j - 1][i - 1] * f[i - j]) % 1000000007;
    }
  }
  cin >> m >> n;
  for (int i = 0; i < (n); ++i) {
    string s;
    cin >> s;
    for (int j = 0; j < (m); ++j) a[i][j] = s[j] - '0';
  }
  memset(flag, 0, sizeof(flag));
  long long res = 1;
  for (int i = 0; i < (m); ++i)
    if (!flag[i]) {
      flag[i] = 1;
      int num = 1;
      for (int j = i + 1; j < m; j++) {
        bool ok = true;
        for (int t = 0; t < (n); ++t)
          if (a[t][i] != a[t][j]) {
            ok = false;
            break;
          }
        if (ok) {
          flag[j] = 1;
          num++;
        }
      }
      res = (res * f[num]) % 1000000007;
    }
  cout << res << endl;
  return 0;
}
