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 M = 1e9 + 7;
const double EPS = 1e-10;
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
const long long LINF = 1e18;
const int N = 1010;
const long long MOD = 1e9 + 7;
int n, m;
long long S[N][N], bellNumber[N];
map<long long, int> mp;
long long st[N];
char s[N];
void pre() {
  S[0][0] = 1;
  for (int i = 0; i <= m; i++) {
    for (int j = 0; j <= i; j++) {
      bellNumber[i] = (bellNumber[i] + S[i][j]) % MOD;
      S[i + 1][j] = (S[i + 1][j] + j * S[i][j]) % MOD;
      S[i + 1][j + 1] = (S[i + 1][j + 1] + S[i][j]);
    }
  }
}
int main() {
  cin >> m >> n;
  pre();
  for (int i = 0; i < n; i++) {
    scanf("%s", s);
    for (int j = 0; j < m; j++) st[j] = st[j] * 2 + (s[j] - '0');
  }
  for (int i = 0; i < m; i++) mp[st[i]]++;
  long long ans = 1;
  for (auto it : mp) ans = (ans * bellNumber[it.second]) % MOD;
  cout << ans << endl;
  return 0;
}
