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 P = 1e9 + 7;
const int MAXM = 1111;
const int MAXN = 55;
void add(int &a, int b) {
  a += b;
  if (a >= P) {
    a -= P;
  }
}
void mul(int &a, int b) {
  long long c = (long long)a * b;
  c %= P;
  a = c;
}
bitset<MAXM> b[55];
int c[MAXM][MAXM];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int m, n;
  cin >> m >> n;
  for (int i = 0; i < n; ++i) {
    string s;
    cin >> s;
    for (int j = 0; j < m; ++j) {
      b[i][j] = (s[j] == '1');
    }
  }
  vector<int> used(m);
  vector<int> cnt;
  for (int j = 0; j < m; ++j) {
    if (used[j]) {
      continue;
    }
    int k = -1;
    for (int i = 0; i < n; ++i) {
      if (b[i][j] == 1) {
        k = i;
        break;
      }
    }
    if (k == -1) {
      continue;
      ;
    }
    for (int i = 0; i < n; ++i) {
      if (i == k) {
        continue;
      }
      if (b[i][j] == 1) {
        b[i] ^= b[k];
      }
    }
    int cur = 0;
    for (int l = 0; l < m; ++l) {
      int s = 0;
      for (int i = 0; i < n; ++i) {
        s += b[i][l];
      }
      if (s == 1 && b[k][l]) {
        ++cur;
        used[l] = 1;
      }
    }
    cnt.push_back(cur);
  }
  int s = m;
  for (int i = 0; i < (int)(cnt).size(); ++i) {
    s -= cnt[i];
  }
  cnt.push_back(s);
  for (int i = 0; i <= m; ++i) {
    c[i][0] = 1;
    for (int j = 1; j <= i; ++j) {
      c[i][j] = c[i - 1][j];
      add(c[i][j], c[i - 1][j - 1]);
    }
  }
  vector<int> f(m + 1);
  f[0] = 1;
  for (int i = 1; i <= m; ++i) {
    for (int k = 1; k <= i; ++k) {
      int cur = f[i - k];
      mul(cur, c[i - 1][k - 1]);
      add(f[i], cur);
    }
  }
  int res = 1;
  for (int i = 0; i < (int)(cnt).size(); ++i) {
    mul(res, f[cnt[i]]);
  }
  cout << res << endl;
}
