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;
int m, n;
char bit[55][1005];
long long fact[1005], fact_i[1005];
long long pow(long long a, long long n) {
  if (n == 0) return 1;
  if (n % 2) {
    return ((a % 1000000007) * (pow(a, n - 1) % 1000000007)) % 1000000007;
  } else {
    return pow((a * a) % 1000000007, n / 2) % 1000000007;
  }
}
void make_fact() {
  long long val = 1;
  fact[0] = 1;
  for (int i = 1; i < 1005; i++) {
    val *= i;
    val %= 1000000007;
    fact[i] = val;
  }
  for (int i = 0; i < 1005; i++) {
    fact_i[i] = pow(fact[i], 1000000007 - 2);
  }
}
long long comb(long long n, long long k) {
  long long ret = 1;
  ret *= fact[n];
  ret *= fact_i[k], ret %= 1000000007;
  ret *= fact_i[n - k], ret %= 1000000007;
  return ret;
}
int parent[1005];
void init() {
  for (int i = 0; i < m; i++) parent[i] = i;
}
int root(int i) {
  if (parent[i] == i) return i;
  return parent[i] = root(parent[i]);
}
bool same(int i, int j) { return root(i) == root(j); }
void unite(int i, int j) {
  int root_i = root(i), root_j = root(j);
  if (root_i == root_j) return;
  parent[root_i] = root_j;
}
long long dp[1005];
int cnt[1005];
int main(void) {
  cin >> m >> n;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      cin >> bit[i][j];
    }
  }
  init();
  make_fact();
  for (int i = 0; i < m; i++) {
    for (int j = 0; j < m; j++) {
      if (i >= j) continue;
      for (int k = 0; k < n; k++) {
        if (bit[k][i] != bit[k][j]) goto pass;
      }
      unite(i, j);
    pass:;
    }
  }
  dp[0] = 1;
  for (int i = 1; i < 1005; i++) {
    for (int j = 1; j <= i; j++) {
      dp[i] += (comb(i - 1, j - 1) * dp[i - j]) % 1000000007;
      dp[i] %= 1000000007;
    }
  }
  long long ans = 1;
  for (int i = 0; i < m; i++) root(i);
  for (int i = 0; i < m; i++) {
    cnt[parent[i]]++;
  }
  for (int i = 0; i < m; i++) {
    if (cnt[i]) {
      ans *= dp[cnt[i]];
      ans %= 1000000007;
    }
  }
  cout << ans << endl;
  return 0;
}
