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 maxm = 1000, maxn = 50, mod = 1e9 + 7;
inline int Add(int x) { return x >= mod ? x - mod : x; }
inline void Add(int &x, int y) {
  x += y;
  if (x >= mod) x -= mod;
}
inline int Mul(int x, int y) { return (long long)x * y % mod; }
int n, m, bio[maxm + 1][maxm + 1], dp[maxm + 1];
char s[maxn][maxm + 1];
map<unsigned long long, int> size;
int main(void) {
  scanf("%d%d", &m, &n);
  for (int i = 0; i < n; ++i) {
    scanf("%s", s[i]);
  }
  for (int i = 0; i <= m; ++i) {
    bio[i][0] = bio[i][i] = 1;
    for (int j = 1; j < i; ++j) {
      bio[i][j] = Add(bio[i - 1][j - 1] + bio[i - 1][j]);
    }
  }
  dp[0] = dp[1] = 1;
  for (int i = 2; i <= m; ++i) {
    dp[i] = 0;
    for (int j = 1; j <= i; ++j) {
      Add(dp[i], Mul(bio[i - 1][j - 1], dp[i - j]));
    }
  }
  for (int i = 0; i < m; ++i) {
    unsigned long long b = 0;
    for (int j = 0; j < n; ++j) {
      b |= (long long)(s[j][i] - '0') << j;
    }
    ++size[b];
  }
  int ans = 1;
  for (auto x : size) {
    ans = Mul(ans, dp[x.second]);
  }
  printf("%d\n", ans);
  return 0;
}
