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 long long mod = 1000000007;
long long powmod(long long a, long long b) {
  long long res = 1;
  a %= mod;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
int n, m, rk;
long long g[1010], comb[1010][1010], bel[1010];
char s[1010];
map<long long, int> hs;
int main() {
  scanf("%d%d", &m, &n);
  for (int i = 0; i < n; i++) {
    scanf("%s", s);
    for (int j = 0; j < m; j++) g[j] = g[j] * 2 + (s[j] - '0');
  }
  for (int j = 0; j < m; j++) hs[g[j]]++;
  long long ret = 1;
  bel[0] = 1;
  comb[0][0] = 1;
  for (int i = 0; i < m + 1; i++) {
    comb[i][0] = 1;
    for (int j = 1; j < i + 1; j++)
      comb[i][j] = (comb[i - 1][j - 1] + comb[i - 1][j]) % mod;
    for (int j = 0; j < i + 1; j++)
      bel[i + 1] = (bel[i + 1] + comb[i][j] * bel[j]) % mod;
  }
  for (auto p : hs) ret = ret * bel[p.second] % mod;
  printf("%lld\n", ret);
}
