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 MAXM = 1005, MAXN = 55, MOD = 1000000007;
int bell[MAXM][MAXM];
void Init_Bell() {
  bell[1][1] = 1;
  for (int i = 2; i < MAXM; i++) {
    bell[i][1] = bell[i - 1][i - 1];
    for (int j = 2; j <= i; j++)
      bell[i][j] = (bell[i][j - 1] + bell[i - 1][j - 1]) % MOD;
  }
}
int n, m;
bitset<MAXM> A[MAXN * 2];
int v[MAXM], tot;
bool vis[MAXM];
void Input() {
  scanf("%d%d", &m, &n);
  for (int i = 1, x; i <= n; i++)
    for (int j = 0; j < m; j++) {
      scanf("%1d", &x);
      A[i][j] = x;
    }
  for (int i = 1; i <= n; i++) A[n + i] = ~A[i];
}
void Make_Partition() {
  bitset<MAXM> t;
  for (int i = 0; i < m; i++) {
    if (vis[i]) continue;
    t.set();
    for (int j = 1; j <= 2 * n; j++)
      if (A[j][i]) t &= A[j];
    int cnt = 0;
    for (int j = 0; j < m; j++)
      if (t[j]) {
        vis[j] = true;
        cnt++;
      }
    v[++tot] = cnt;
  }
}
int solve() {
  int ans = 1;
  for (int i = 1; i <= tot; i++) ans = (1LL * ans * bell[v[i]][v[i]]) % MOD;
  return ans;
}
int main() {
  Input();
  Init_Bell();
  Make_Partition();
  printf("%d\n", solve());
  return 0;
}
