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 int M = 1000000007;
int n, m, i, j;
long long f[1005][1005], ans = 1, a[1005];
map<long long, int> mp;
char c[55][1005];
int main() {
  scanf("%d %d", &m, &n);
  f[0][0] = 1;
  for (i = 1; i <= 1000; ++i)
    for (j = 1; j <= i; ++j) f[i][j] = (f[i - 1][j] * j + f[i - 1][j - 1]) % M;
  for (i = 1; i <= 1000; ++i)
    for (j = 1; j <= i; ++j) a[i] = (a[i] + f[i][j]) % M;
  for (i = 1; i <= n; ++i) scanf("%s", c[i] + 1);
  for (i = 1; i <= m; ++i) {
    long long s = 0;
    for (j = 1; j <= n; ++j) s |= 1ll * (c[j][i] - '0') << j;
    ++mp[s];
  }
  for (auto it : mp) ans = ans * a[it.second] % M;
  cout << (ans % M + M) % M;
}
