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;
bitset<1010> b[1010], ans, T;
int m, n, B[1010], fac[1010], rfac[1010], ri[1010];
const int mod = 1e9 + 7;
inline int C(int n, int m) {
  return 1ll * rfac[m] * rfac[n - m] % mod * fac[n] % mod;
}
int main() {
  fac[0] = rfac[0] = ri[1] = 1;
  for (int i = 1; i < 1010; i++) {
    fac[i] = 1ll * i * fac[i - 1] % mod;
    if (i > 1) ri[i] = 1ll * (mod - mod / i) * ri[mod % i] % mod;
    rfac[i] = 1ll * ri[i] * rfac[i - 1] % mod;
  }
  B[0] = 1;
  for (int i = 1; i < 1010; i++)
    for (int j = 0; j < i; j++) B[i] = (B[i] + 1ll * C(i - 1, j) * B[j]) % mod;
  scanf("%d %d", &m, &n);
  for (int i = 0; i < n; i++) cin >> b[i];
  for (int i = 0; i < m; i++) T[i] = 1;
  int rlt = 1;
  for (int i = 0; i < m; i++) {
    if (ans[i]) continue;
    bitset<1010> x = T;
    for (int j = 0; j < n; j++)
      if (b[j][i])
        x &= b[j];
      else
        x &= (~b[j] & T);
    rlt = 1ll * rlt * B[x.count()] % mod;
    ans |= x;
  }
  printf("%d\n", rlt);
  return 0;
}
