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 long long MAX = 1001;
const long long MOD = 1e9 + 7;
long long dp[MAX][MAX];
long long bell[MAX];
void combinationDp() {
  for (int i = 0; i < MAX; i++) dp[i][0] = 1;
  for (int i = 1; i < MAX; i++) {
    for (int j = 1; j < MAX; j++) {
      if (i >= j)
        dp[i][j] = (dp[i - 1][j - 1] % MOD + dp[i - 1][j] % MOD) % MOD;
      else
        dp[i][j] = 0;
    }
  }
  return;
}
void bellFunc() {
  bell[0] = 1;
  for (int i = 1; i < MAX; i++) {
    for (int j = 0; j < i; j++) {
      bell[i] += (bell[j] % MOD * dp[i - 1][j] % MOD) % MOD;
    }
    bell[i] %= MOD;
  }
}
int main() {
  long long m, n;
  cin >> m >> n;
  combinationDp();
  bellFunc();
  vector<string> tmp(n);
  for (int i = 0; i < n; i++) cin >> tmp[i];
  unordered_map<string, long long> mp;
  for (int i = 0; i < m; i++) {
    string s;
    for (int j = 0; j < n; j++) {
      s.push_back(tmp[j][i]);
    }
    mp[s]++;
  }
  long long res = 1;
  for (auto c : mp) {
    res *= bell[c.second] % MOD;
    res %= MOD;
  }
  cout << res;
}
