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 N = 1000 + 5;
const int MOD = (int)1e9 + 7;
int m, n;
char s[N][N];
long long binom[N][N];
long long dp[N];
void solve() {
  binom[0][0] = 1;
  for (int i = 1; i < N; ++i) {
    binom[i][0] = 1;
    for (int j = 1; j < N; ++j) {
      binom[i][j] = (binom[i - 1][j] + binom[i - 1][j - 1]) % MOD;
    }
  }
  dp[0] = 1;
  dp[1] = 1;
  for (int i = 2; i < N; ++i) {
    dp[i] = 0;
    for (int j = 0; j < i; ++j) {
      dp[i] += dp[i - 1 - j] * binom[i - 1][j] % MOD;
    }
    dp[i] %= MOD;
  }
  cin >> m >> n;
  for (int i = 0; i < n; ++i) {
    scanf("%s", s[i]);
    for (int j = 0; j < m; ++j) s[i][j] -= '0';
  }
  map<long long, int> freq;
  for (int j = 0; j < m; ++j) {
    long long x = 0;
    for (int i = 0; i < n; ++i) {
      if (s[i][j]) x += 1ll << i;
    }
    freq[x]++;
  }
  long long ret = 1;
  for (auto e : freq) {
    long long x = e.second;
    ret = ret * dp[x] % MOD;
  }
  cout << ret << endl;
}
int main() {
  solve();
  return 0;
}
