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;
long long n, m, f[1010][1010], dp[1010], p[1010], ans = 1, x;
map<long long, long long> mp;
int main() {
  int i;
  for (scanf("%lld%lld", &n, &m); m--;)
    for (i = 0; i < n; ++i) scanf("%1lld", &x), p[i] |= x << m;
  for (i = 0; i < n; ++i) ++mp[p[i]];
  for (i = 0; i <= n; ++i)
    for (int j = 0; j <= i; ++j)
      f[i][j] = j ? (f[i - 1][j - 1] + f[i - 1][j]) % mod : 1;
  for (dp[0] = dp[1] = 1, i = 2; i <= n; ++i)
    for (int j = 0; j < i; j++) dp[i] = (dp[i] + f[i - 1][j] * dp[j]) % mod;
  for (map<long long, long long>::iterator t = mp.begin(); t != mp.end(); ++t)
    ans = (ans * dp[t->second]) % mod;
  printf("%lld", ans);
}
