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 N, M, c[1009][1009], bell[1009];
char sir[60][1009];
const int mod = 1e9 + 7;
long long v[1009];
int add(int x, int y) {
  int ans = x + y;
  if (ans >= mod) ans -= mod;
  return ans;
}
int subtract(int x, int y) {
  if (x >= y) return x - y;
  return x - y + mod;
}
int mul(int x, int y) { return 1LL * x * y % mod; }
void adto(int &x, int y) {
  x += y;
  if (x >= mod) x -= mod;
}
int main() {
  scanf("%d %d\n", &M, &N);
  c[0][0] = 1;
  for (int i = 1; i <= M; i++) {
    c[i][0] = 1;
    for (int j = 1; j <= i; j++) c[i][j] = add(c[i - 1][j - 1], c[i - 1][j]);
  }
  bell[0] = bell[1] = 1;
  for (int i = 1; i < M; i++)
    for (int j = 0; j <= i; j++) adto(bell[i + 1], mul(c[i][j], bell[j]));
  for (int i = 1; i <= N; i++) {
    gets(sir[i] + 1);
    for (int j = 1; j <= M; j++) sir[i][j] -= '0';
  }
  for (int j = 1; j <= M; j++) {
    v[j] = 0;
    for (int i = 1; i <= N; i++)
      if (sir[i][j]) v[j] |= 1LL << i;
  }
  sort(v + 1, v + M + 1);
  int ans = 1;
  for (int i = 1; i <= M; i++) {
    int j = i;
    for (; j <= M; j++)
      if (v[j] != v[i]) break;
    j--;
    ans = mul(ans, bell[j - i + 1]);
    i = j;
  }
  printf("%d\n", ans);
  return 0;
}
