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;
int F[1005][1005];
long long vd[1005];
int T[1005], tot = 0;
int n, m;
long long dp[1005][1005];
char cc[10005];
long long P2[10005];
int main() {
  scanf("%d%d", &m, &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%s", cc + 1);
    for (int j = 1; j <= m; ++j) {
      for (int k = j + 1; k <= m; ++k) {
        if (cc[j] != cc[k]) F[j][k] = F[k][j] = 1;
      }
    }
  }
  for (int i = 1; i <= m; ++i) {
    if (vd[i]) continue;
    ++T[++tot];
    for (int j = i + 1; j <= m; ++j) {
      if (!F[i][j]) ++T[tot], vd[j] = 1;
    }
  }
  memset(dp, 0, sizeof dp);
  memset(vd, 0, sizeof vd);
  dp[1][1] = 1;
  for (int i = 2; i <= m; ++i) {
    for (int j = 1; j <= i; ++j) {
      dp[i][j] += dp[i - 1][j - 1] + dp[i - 1][j] * j % 1000000007;
      dp[i][j] %= 1000000007;
    }
  }
  for (int i = 1; i <= m; ++i)
    for (int j = 1; j <= i; ++j) vd[i] += dp[i][j], vd[i] %= 1000000007;
  long long res = 1;
  for (int i = 1; i <= tot; ++i) res = res * vd[T[i]] % 1000000007;
  cout << res;
  return 0;
}
