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 = 1000000007;
long long v[1010];
char s[1010];
long long dp[1010][1010];
int cnt;
long long work(int n) {
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= i; j++) dp[i][j] = 0;
  dp[0][0] = 1;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= i; j++)
      dp[i][j] = (dp[i - 1][j - 1] + dp[i - 1][j] * j) % mod;
  long long ret = 0;
  for (int i = 1; i <= n; i++) ret += dp[n][i];
  return ret % mod;
}
int main() {
  int m, n;
  long long ans = 1;
  scanf("%d%d", &m, &n);
  memset(dp, 0, sizeof(dp));
  memset(v, 0, sizeof(v));
  for (int i = 1; i <= n; i++) {
    scanf("%s", s + 1);
    for (int j = 1; j <= m; j++) v[j] = (v[j] << 1) | (s[j] - '0');
  }
  sort(v + 1, v + 1 + m);
  int cnt = 0;
  v[0] = v[1];
  v[m + 1] = -1;
  m++;
  for (int i = 1; i <= m; i++)
    if (v[i] == v[i - 1])
      cnt++;
    else {
      (ans *= work(cnt)) %= mod;
      cnt = 1;
    }
  printf("%I64d", ans);
  return 0;
}
