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>
#pragma GCC optimize("Ofast,unroll-loops,no-stack-protector,fast-math")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma comment(linker, "/STACK:16777216")
using namespace std;
using matrix = vector<vector<long long>>;
void FAST_IO() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
const long long mod = 1e9 + 7;
const int MAXN = 1050;
bool eq[MAXN][MAXN], used[MAXN];
long long dp[MAXN][MAXN], B[MAXN];
signed main() {
  FAST_IO();
  memset(eq, true, sizeof(eq));
  B[0] = dp[0][0] = 1;
  for (int i = 1; i < MAXN; ++i) {
    for (int j = 1; j <= i; ++j) {
      dp[i][j] = (dp[i - 1][j - 1] + j * dp[i - 1][j]) % mod;
      B[i] = (B[i] + dp[i][j]) % mod;
    }
  }
  int m, n;
  cin >> m >> n;
  while (n--) {
    string s;
    cin >> s;
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < m; ++j) {
        if (s[i] != s[j]) {
          eq[i][j] = false;
        }
      }
    }
  }
  long long ans = 1;
  for (int i = 0; i < m; ++i) {
    if (used[i]) {
      continue;
    }
    int cnt = 0;
    for (int j = i; j < m; ++j) {
      if (eq[i][j]) {
        cnt++;
        used[j] = true;
      }
    }
    ans = (ans * B[cnt]) % mod;
  }
  cout << ans;
}
