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 mod = 1e9 + 7;
int n, m, f[100005], fr[100005], infr[100005];
long long a[100005];
inline int addmod(int x) { return x >= mod ? x - mod : x; }
inline int submod(int x) { return x < 0 ? x + mod : x; }
int getans(int x, int y) {
  return 1ll * fr[x] * infr[y] % mod * infr[x - y] % mod;
}
int fpow(int x, int y) {
  int ans = 1;
  while (y) {
    if (y & 1) ans = 1ll * ans * x % mod;
    x = 1ll * x * x % mod;
    y /= 2;
  }
  return ans;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++)
    for (int j = 1; j <= n; j++) {
      char ch;
      scanf(" %c", &ch);
      a[j] |= ((long long)(ch - '0') << (long long)(i - 1));
    }
  sort(a + 1, a + n + 1);
  fr[0] = infr[0] = 1;
  for (int i = 1; i <= n; i++) {
    fr[i] = 1ll * i * fr[i - 1] % mod;
    infr[i] = fpow(fr[i], mod - 2);
  }
  f[0] = 1;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= i; j++)
      f[i] = addmod(f[i] + 1ll * getans(i - 1, j - 1) * f[i - j] % mod);
  int ans = 1, ct = 1;
  long long val = a[1];
  a[n + 1] = -1;
  for (int i = 2; i <= n + 1; i++)
    if (val == a[i])
      ct++;
    else {
      ans = 1ll * ans * f[ct] % mod;
      ct = 1;
      val = a[i];
    }
  printf("%d", ans);
  return 0;
}
