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;
int m, n;
long long s[1005];
long long dp[1005][1005];
long long tdp[1005];
void init() {
  scanf("%d%d", &m, &n);
  for (int i = 0; i < n; i++) {
    char st[1005];
    scanf("%s", st);
    for (int j = 0; j < m; j++) {
      if (st[j] == '1') s[j] += (1ll << i);
    }
  }
  dp[1][1] = 1;
  for (int i = 1; i <= 1000; i++) {
    for (int j = 1; j <= i; j++) {
      tdp[i] = (tdp[i] + dp[i][j]) % mod;
      dp[i + 1][j + 1] = (dp[i + 1][j + 1] + dp[i][j]) % mod;
      dp[i + 1][j] = (dp[i + 1][j] + dp[i][j] * j % mod) % mod;
    }
  }
}
map<long long, int> mp;
void process() {
  for (int i = 0; i < m; i++) {
    mp[s[i]]++;
  }
  int ans = 1;
  for (pair<int, int> pr : mp) {
    ans = (long long)ans * tdp[pr.second] % mod;
  }
  printf("%d\n", ans);
}
int main() {
  init();
  process();
  return 0;
}
