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 mo = 1e9 + 7;
const int N = 1000;
long long fac[N + 10], inv[N + 10], f[N + 10];
int m, n;
string s[N + 10];
char ss[N + 10];
long long mul(long long x, long long y) {
  if (y == 0) return 1;
  long long now = mul(x, y / 2);
  now = now * now % mo;
  if (y % 2) return now * x % mo;
  return now;
}
void prework() {
  fac[0] = 1;
  for (int i = 1; i <= N; i++) fac[i] = fac[i - 1] * i % mo;
  inv[N] = mul(fac[N], mo - 2);
  for (int i = N - 1; i >= 0; i--) inv[i] = inv[i + 1] * (i + 1) % mo;
}
long long C(long long x, long long y) {
  return fac[x] * inv[y] % mo * inv[x - y] % mo;
}
int main() {
  prework();
  f[0] = f[1] = 1;
  for (int i = 2; i <= N; i++)
    for (int j = 0; j < i; j++)
      f[i] = (f[i] + C(i - 1, j) * f[i - 1 - j] % mo) % mo;
  scanf("%d%d", &m, &n);
  for (int i = 1; i <= m; i++) s[i] = "";
  for (int i = 1; i <= n; i++) {
    scanf("%s", ss);
    for (int j = 0; j < m; j++) s[j + 1] += ss[j];
  }
  sort(s + 1, s + m + 1);
  long long ans = 1;
  for (int i = 1; i <= m;) {
    int j = i;
    while (j <= m && s[j] == s[i]) j++;
    ans = ans * f[j - i] % mo;
    i = j;
  }
  printf("%lld\n", ans);
  return 0;
}
