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;
template <typename T, typename U>
inline void smin(T &a, U b) {
  if (a > b) a = b;
}
template <typename T, typename U>
inline void smax(T &a, U b) {
  if (a < b) a = b;
}
template <class T>
inline void gn(T &first) {
  char c, sg = 0;
  while (c = getchar(), (c > '9' || c < '0') && c != '-')
    ;
  for ((c == '-' ? sg = 1, c = getchar() : 0), first = 0; c >= '0' && c <= '9';
       c = getchar())
    first = (first << 1) + (first << 3) + c - '0';
  if (sg) first = -first;
}
template <class T, class T1>
inline void gn(T &first, T1 &second) {
  gn(first);
  gn(second);
}
template <class T, class T1, class T2>
inline void gn(T &first, T1 &second, T2 &z) {
  gn(first);
  gn(second);
  gn(z);
}
template <class T>
inline void print(T first) {
  if (first < 0) {
    putchar('-');
    return print(-first);
  }
  if (first < 10) {
    putchar('0' + first);
    return;
  }
  print(first / 10);
  putchar(first % 10 + '0');
}
template <class T>
inline void printsp(T first) {
  print(first);
  putchar(' ');
}
template <class T>
inline void println(T first) {
  print(first);
  putchar('\n');
}
int power(int a, int b, int m, int ans = 1) {
  for (; b; b >>= 1, a = 1LL * a * a % m)
    if (b & 1) ans = 1LL * ans * a % m;
  return ans;
}
int fac[1010], rev[1010];
int dp[1010][1010], a[1010];
char str[1010][1010];
void pre() {
  fac[0] = rev[0] = 1;
  for (int i = 1; i < 1010; i++) {
    fac[i] = 1LL * fac[i - 1] * i % 1000000007;
    rev[i] = power(fac[i], 1000000007 - 2, 1000000007);
  }
  for (int i = 0; i < 1010; i++) dp[0][i] = 1;
  for (int i = 1; i < 1010; i++) {
    for (int j = 1; j < 1010; j++) {
      for (int k = 0, p = 1; k * j <= i;
           k++, p = 1LL * p * rev[j] % 1000000007) {
        dp[i][j] = (dp[i][j] + 1LL * dp[i - k * j][j - 1] * p % 1000000007 *
                                   rev[k] % 1000000007) %
                   1000000007;
      }
    }
  }
}
int main() {
  pre();
  int n, m;
  cin >> m >> n;
  for (int i = 0; i < n; i++) scanf("%s", str[i]);
  for (int i = 0; i < m; i++) {
    for (int j = 0; j < n; j++) {
      a[i] = (a[i] << 1) + (str[j][i] - '0');
    }
  }
  long long ans = 1;
  sort(a, a + m);
  for (int i = 0, j; i < m; i = j) {
    for (j = i; j < m; j++)
      if (a[i] != a[j]) break;
    ans = ans * dp[j - i][j - i] % 1000000007 * fac[j - i] % 1000000007;
  }
  println(ans);
  return 0;
}
