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 MOD = (1e9) + 7;
int M, N;
string S[55];
long long B[1005];
long long C[1005][1005];
long long V[1005];
int add(int a, int b) {
  a += b;
  if (a >= MOD) {
    a -= MOD;
  }
  return a;
}
int mult(int a, int b) { return 1LL * a * b % MOD; }
int main() {
  cin >> M >> N;
  for (int i = 1; i <= N; i++) {
    cin >> S[i];
    S[i] = " " + S[i];
  }
  for (int i = 0; i <= M; i++) {
    C[i][0] = C[i][i] = 1;
    for (int j = 1; j < i; j++) {
      C[i][j] = add(C[i - 1][j], C[i - 1][j - 1]);
    }
  }
  B[0] = B[1] = 1;
  for (int i = 2; i <= M; i++) {
    for (int j = 0; j < i; j++) {
      B[i] = add(B[i], mult(B[i - j - 1], C[i - 1][j]));
    }
  }
  for (int i = 1; i <= M; i++) {
    for (int j = 1; j <= N; j++) {
      if (S[j][i] == '1') {
        V[i] |= 1LL << (j - 1);
      }
    }
  }
  sort(V + 1, V + 1 + M);
  V[0] = V[M + 1] = -1;
  int nr = 0;
  int rez = 1;
  for (int i = 1; i <= M + 1; i++) {
    if (V[i] == V[i - 1]) {
      nr++;
    } else {
      rez = mult(rez, B[nr]);
      nr = 1;
    }
  }
  cout << rez;
  return 0;
}
