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 n, m;
long long hsh[2333], C[2333][2333], B[2333];
map<long long, int> cnt;
char str[2333];
const int MOD = 1e9 + 7;
int main() {
  for (int i = 0; i <= 2000; ++i) {
    C[i][0] = 1;
    for (int j = 1; j <= i; ++j)
      C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % MOD;
  }
  B[0] = B[1] = 1;
  for (int i = 1; i <= 2000; ++i)
    for (int j = 0; j <= i; ++j) (B[i + 1] += C[i][j] * B[j]) %= MOD;
  scanf("%d%d", &m, &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%s", str);
    for (int j = 0; str[j]; ++j)
      if (str[j] == '1') hsh[j] ^= 1LL << i;
  }
  for (int i = 0; i < m; ++i) cnt[hsh[i]]++;
  long long ans = 1;
  for (auto s : cnt) ans = ans * B[s.second] % MOD;
  ans = (ans % MOD + MOD) % MOD;
  printf("%d\n", int(ans));
}
