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 P = 1e9 + 7, INF = 0x3f3f3f3f;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long qpow(long long a, long long n) {
  long long r = 1 % P;
  for (a %= P; n; a = a * a % P, n >>= 1)
    if (n & 1) r = r * a % P;
  return r;
}
long long inv(long long first) {
  return first <= 1 ? 1 : inv(P % first) * (P - P / first) % P;
}
inline int rd() {
  int first = 0;
  char p = getchar();
  while (p < '0' || p > '9') p = getchar();
  while (p >= '0' && p <= '9') first = first * 10 + p - '0', p = getchar();
  return first;
}
const int N = 1e3 + 10;
int n, m, vis[N], a[N][N], b[N];
char s[N][N];
map<long long, int> f;
int main() {
  scanf("%d%d", &m, &n);
  for (int i = 1; i <= n; ++i) scanf("%s", s[i] + 1);
  for (int i = 1; i <= m; ++i) {
    long long first = 0;
    for (int j = 1; j <= n; ++j) first = first << 1 ^ (s[j][i] == '1');
    ++f[first];
  }
  a[0][0] = 1;
  for (int i = 1; i <= m; ++i)
    for (int j = 1; j <= i; ++j) {
      a[i][j] = (a[i - 1][j - 1] + (long long)j * a[i - 1][j]) % P;
      b[i] = (b[i] + a[i][j]) % P;
    }
  long long ans = 1;
  for (auto t : f) ans = ans * b[t.second] % P;
  printf("%lld\n", ans);
}
