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;
const int maxn = 1e5 + 5;
int add(int x, int y) {
  x += y;
  if (x >= mod) x -= mod;
  return x;
}
int mul(int x, int y) {
  long long z = 1LL * x * y;
  return z - z / mod * mod;
}
int powt(int a, int b) {
  int r = 1;
  while (b) {
    if (b & 1) r = mul(r, a);
    a = mul(a, a);
    b >>= 1;
  }
  return r;
}
char s[1005];
char f[1005][55];
int g[1005];
int c[1005], b[1005];
int C(int x, int y) { return mul(c[x], powt(mul(c[y], c[x - y]), mod - 2)); }
int main() {
  int m, n, i, j;
  scanf("%d%d", &m, &n);
  c[0] = 1;
  for (i = 1; i <= m; i++) c[i] = mul(c[i - 1], i);
  b[0] = 1;
  for (i = 1; i <= m; i++) {
    for (j = 0; j < i; j++) b[i] = add(b[i], mul(b[j], C(i - 1, j)));
  }
  for (i = 0; i < n; i++) {
    scanf(" %s", s);
    for (j = 0; j < m; j++) f[j][i] = s[j];
  }
  for (i = 0; i < m; i++) f[i][n] = 0;
  for (i = 0; i < m; i++) {
    for (j = 0; j < i; j++) {
      if (strcmp(f[i], f[j]) == 0) break;
    }
    g[j]++;
  }
  int ans = 1;
  for (i = 0; i < m; i++) {
    ans = mul(ans, b[g[i]]);
  }
  printf("%d\n", ans);
  return 0;
}
