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;
#pragma comment(linker, "/STACK:102400000,102400000")
const long long MOD = 1e9 + 7;
const int N = 1007;
int m, n;
char s[N];
long long a[N];
long long b[N];
long long dp[N][N];
int main() {
  scanf("%d%d", &m, &n);
  for (int i = int(1); i <= int(n); ++i) {
    scanf("%s", s + 1);
    for (int i = int(1); i <= int(m); ++i) {
      a[i] <<= 1;
      a[i] += 1LL * (s[i] - '0');
    }
  }
  sort(a + 1, a + 1 + m);
  dp[0][0] = 1;
  for (int i = int(1); i <= int(m); ++i) {
    for (int j = int(1); j <= int(i); ++j) {
      dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j] * j % MOD;
      dp[i][j] %= MOD;
      (b[i] += dp[i][j]) %= MOD;
    }
  }
  long long ans = 1;
  int i = 1, cnt = 1;
  while (i <= m) {
    while (i <= m && a[i] == a[i + 1]) cnt++, i++;
    (ans *= b[cnt]) %= MOD;
    cnt = 1;
    i++;
  }
  cout << ans << endl;
}
