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>
#pragma comment(linker, "/STACK:536870912")
#pragma warning(disable : 4996)
using namespace std;
const int M = 1005, N = 55, mod = 1e9 + 7;
bitset<M> c[N];
long long int dp[M][M];
bool us[M];
int main() {
  dp[0][0] = 1;
  for (int i = 1; i < M; i++) {
    for (int j = 1; j < M; j++) {
      dp[i][j] = (dp[i - 1][j - 1] + dp[i - 1][j] * j) % mod;
    }
  }
  int m, n;
  scanf("%d%d", &m, &n);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      char a;
      scanf(" %c", &a);
      c[i][j] = a - '0';
    }
  }
  long long int ans = 1;
  for (int i = 0; i < m; i++) {
    if (us[i]) continue;
    bitset<M> r;
    for (int j = 0; j < m; j++) r[j] = 1;
    for (int j = 0; j < n; j++) {
      r &= c[j][i] ? c[j] : ~c[j];
    }
    for (int j = 0; j < m; j++) {
      if (r[j]) us[j] = 1;
    }
    long long int t = 0, ct = r.count();
    for (int j = 0; j <= ct; j++) t = (t + dp[ct][j]) % mod;
    ans = ans * t % mod;
  }
  printf("%lld", ans);
  return 0;
}
