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 = 1e9 + 7;
int par[1];
int find(int x) { return x == par[x] ? x : par[x] = find(par[x]); }
void join(int x, int y) { par[find(x)] = y; }
long long mod_expo(long long a, long long b, long long mod) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return ans;
}
long long gcd(long long a, long long b, long long& x, long long& y) {
  if (!a) {
    x = 0;
    y = 1;
    return b;
  }
  long long x1, y1;
  long long d = gcd(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  return d;
}
const int N = 1005;
long long c[N][N], b[N], a[N];
char s[N];
map<long long, int> mp;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  int i, j, k;
  cin >> m >> n;
  for (i = 0; i < n; ++i) {
    cin >> s;
    for (j = 0; j < m; ++j) {
      a[j] |= ((s[j] - '0' + 0LL) << i);
    }
  }
  for (i = 0; i < m; ++i) {
    ++mp[a[i]];
  }
  for (i = 0; i <= m; ++i) {
    c[i][0] = 1;
  }
  for (i = 1; i <= m; ++i) {
    for (j = 1; j <= i; ++j) {
      c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % MOD;
    }
  }
  b[0] = 1;
  for (i = 1; i <= m; ++i) {
    for (j = 0; j < i; ++j) {
      b[i] = (b[i] + b[j] * c[i - 1][j]) % MOD;
    }
  }
  long long ans = 1;
  for (pair<long long, long long> x : mp) {
    ans = (ans * b[x.second]) % MOD;
  }
  cout << ans << "\n";
  return 0;
}
