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 NSIZ = 1010;
const int MSIZ = 1000010;
const int inf = 1010580540;
const int mxint = 2147483647;
const long long mxll = 9223372036854775807LL;
const long long mod = 1000000007LL;
int n, m, o, re = 0, test;
long long res = 0, dp[NSIZ][NSIZ], sm[NSIZ];
char a[NSIZ][NSIZ], b[NSIZ];
char c[NSIZ];
bool chk[NSIZ];
void prep() {
  dp[1][1] = 1;
  sm[1] = 1;
  for (int i = 2; i < NSIZ; i++) {
    for (int j = 1; j <= i; j++) {
      dp[i][j] = (dp[i - 1][j] * j + dp[i - 1][j - 1]) % mod;
      sm[i] += dp[i][j];
    }
    sm[i] %= mod;
  }
}
int main() {
  int i, j, k, l;
  long long ll = 0, rr = mxll, mid;
  prep();
  scanf("%d %d", &m, &n);
  for (i = 0; i < n; i++) {
    scanf("%s", a[i]);
  }
  res = 1;
  for (i = 0; i < m; i++) {
    l = 1;
    for (j = i + 1; j < m; j++) {
      if (chk[j]) continue;
      for (k = 0; k < n; k++) {
        if (a[k][j] != a[k][i]) break;
      }
      if (k < n) continue;
      chk[j] = 1;
      l++;
    }
    res *= sm[l];
    res %= mod;
  }
  printf("%lld", res);
  return 0;
}
