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;
long long powmod(long long a, long long n) {
  if (n == 0) return 1;
  if (n % 2) return (a * powmod(a, n - 1)) % 1000000007;
  long long c = powmod(a, n / 2);
  return (c * c) % 1000000007;
}
long long inv(long long a) { return powmod(a, 1000000007 - 2); }
long long fact[1100000];
long long invfact[1100000];
long long ncr(long long n, long long r) {
  if (r < 0 || n < 0) return 0;
  if (n < r) return 0;
  long long a = fact[n];
  a = (a * invfact[r]) % 1000000007;
  a = (a * invfact[n - r]) % 1000000007;
  return a;
}
void init() {
  fact[0] = 1;
  invfact[0] = 1;
  long long inv2 = inv(2);
  for (long long i = 1; i < 110000; i++) {
    fact[i] = (i * fact[i - 1]) % 1000000007;
    invfact[i] = inv(fact[i]);
  }
}
long long dp[1100];
int main() {
  init();
  int n, m;
  cin >> m >> n;
  string g[n];
  for (int i = 0; i < n; i++) cin >> g[i];
  for (int i = 0; i < n; i++) {
    if (g[i][0] == '1') {
      for (int j = 0; j < m; j++) {
        g[i][j] = '0' + '1' - g[i][j];
      }
    }
  }
  map<long long, int> f;
  for (int j = 1; j < m; j++) {
    long long c = 0;
    for (int i = 0; i < n; i++) {
      c *= 2;
      c += g[i][j] - '0';
    }
    f[c]++;
  }
  for (int i = 0; i <= 1000; i++) dp[i] = 0;
  dp[0] = 1;
  dp[1] = 1;
  for (int i = 2; i <= 1000; i++) {
    for (int k = 0; k < i; k++) {
      dp[i] += dp[k] * ncr(i - 1, k);
      dp[i] %= 1000000007;
    }
  }
  long long ans = 1;
  for (map<long long, int>::iterator it = f.begin(); it != f.end(); it++) {
    if (it->first > 0) {
      ans = (ans * dp[it->second]) % 1000000007;
    }
  }
  long long ans0 = 0;
  long long num = f[0];
  for (int i = 0; i <= num; i++) {
    ans0 += ncr(num, i) * dp[num - i];
    ans0 %= 1000000007;
  }
  ans = (ans * ans0) % 1000000007;
  cout << ans << endl;
}
