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;
int m, n;
char s[1010];
long long d[1010][1010], num[1010];
long long ans = 1;
vector<bitset<1010> > v;
bitset<1010> b1;
int main(int argc, char const *argv[]) {
  for (int i = 1; i < 1010; ++i) {
    for (int j = 1; j <= i; ++j) {
      if (j == 1)
        d[i][j] = 1;
      else if (j == i)
        d[i][j] = 1;
      else {
        d[i][j] = (d[i - 1][j] * j + d[i - 1][j - 1]) % mod;
      }
      num[i] = (num[i] + d[i][j]) % mod;
    }
  }
  scanf("%d %d", &m, &n);
  for (int i = 0; i < m; ++i) {
    b1.set(i);
  }
  v.push_back(b1);
  for (int i = 1; i <= n; ++i) {
    scanf("%s", s);
    bitset<1010> b;
    for (int j = 0; s[j]; ++j) {
      if (s[j] == '1') b.set(j);
    }
    vector<bitset<1010> > w;
    for (int j = 0; j < v.size(); ++j) {
      bitset<1010> btmp1 = v[j] & b;
      bitset<1010> btmp2 = v[j] ^ btmp1;
      if (btmp1.count() && btmp2.count()) {
        w.push_back(btmp1);
        w.push_back(btmp2);
      } else
        w.push_back(v[j]);
    }
    v = w;
  }
  for (int i = 0; i < v.size(); ++i) {
    int cnt = v[i].count();
    ans = (ans * num[cnt]) % mod;
  }
  printf("%lld\n", ans);
  return 0;
}
