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 = 1000000007;
int dp[1003][1003], b[1003], m, n;
char buf[1003];
bitset<1000> bs[53], all;
inline void add(int &a, int b) {
  a += b;
  if (a >= P) a -= P;
}
inline int mul(int a, int b) { return 1LL * a * b % P; }
int pw(int a, int b) {
  if (b == 0) return 1;
  int res = pw(a, b / 2);
  if (b & 1) return mul(mul(res, res), a);
  return mul(res, res);
}
int inv(int a) {
  assert(a);
  return pw(a, P - 2);
}
int main() {
  scanf("%d %d", &m, &n);
  dp[1][1] = 1;
  for (int i = 1; i <= m; i++) {
    for (int j = 1; j <= i; j++) {
      add(dp[i][j], dp[i - 1][j - 1]);
      add(dp[i][j], mul(j, dp[i - 1][j]));
    }
  }
  for (int i = 1; i <= m; i++) {
    for (int j = 1; j <= i; j++) {
      add(b[i], dp[i][j]);
    }
  }
  for (int i = 0; i < n; i++) {
    scanf("%s", buf);
    bs[i] = bitset<1000>(buf);
  }
  for (int i = 0; i < m; i++) {
    all.set(i);
  }
  vector<int> v;
  bitset<1000> vis;
  for (int i = 0; i < m; i++) {
    if (vis[i]) continue;
    bitset<1000> now = all;
    for (int j = 0; j < n; j++) {
      if (bs[j][i])
        now &= bs[j];
      else
        now &= bs[j] ^ all;
    }
    vis |= now;
    v.push_back((int)now.count());
  }
  int ans = 1;
  for (auto &c : v) {
    ans = mul(ans, b[c]);
  }
  cout << ans << endl;
  return 0;
}
