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 read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int n, m;
map<long long, int> mp;
long long fac[1005], inv[1005], bel[1005];
char s[55][1005];
long long Pow(long long x, long long y) {
  long long ans = 1;
  while (y) {
    if (y & 1) ans = ans * x % 1000000007;
    x = x * x % 1000000007;
    y /= 2;
  }
  return ans;
}
long long get(int x) {
  long long ans = 0;
  for (int i = 1; i <= n; i++) ans = ans * 2 + s[i][x] - '0';
  return ans;
}
long long C(int n, int m) {
  return fac[n] * inv[m] % 1000000007 * inv[n - m] % 1000000007;
}
int main() {
  m = read();
  n = read();
  for (int i = 1; i <= n; i++) scanf("%s", s[i] + 1);
  for (int i = 1; i <= m; i++) mp[get(i)]++;
  fac[0] = 1;
  for (int i = 1; i <= 1000; i++) fac[i] = fac[i - 1] * i % 1000000007;
  inv[1000] = Pow(fac[1000], 1000000007 - 2);
  for (int i = 1000; i >= 1; i--) inv[i - 1] = inv[i] * i % 1000000007;
  bel[0] = 1;
  bel[1] = 1;
  for (int i = 2; i <= m; i++) {
    for (int j = 0; j <= i - 1; j++)
      bel[i] = (bel[i] + C(i - 1, j) * bel[j] % 1000000007) % 1000000007;
  }
  long long ans = 1;
  for (int i = 1; i <= m; i++) {
    long long tmp = get(i);
    ans = ans * bel[mp[tmp]] % 1000000007;
    mp[tmp] = 0;
  }
  printf("%lld\n", ans);
  return 0;
}
