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 long long mod = 1e9 + 7;
int n, m;
char s[5005];
long long state[5005];
long long dp[5005][5005];
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    scanf("%s", s + 1);
    for (int j = 1; j <= n; j++)
      if (s[j] == '1') state[j] |= (1ll << (i - 1));
  }
  dp[0][0] = 1;
  for (int i = 1; i <= n; i++) {
    dp[i][0] = dp[i - 1][i - 1];
    for (int j = 1; j <= n; j++) {
      dp[i][j] = (dp[i - 1][j - 1] + dp[i][j - 1]);
      if (dp[i][j] >= mod) dp[i][j] -= mod;
    }
  }
  sort(state + 1, state + 1 + n);
  long long ans = 1;
  for (int i = 1; i <= n; i++) {
    int j = i;
    for (j = i; state[j] == state[i] && j <= n; j++)
      ;
    ans *= (dp[j - i][0]);
    ans %= mod;
    i = j - 1;
  }
  printf("%I64d\n", ans);
}
