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;
char S[1005][55];
int M, N;
int DP[1005][1006];
int C[1005][1005];
int Sum[1005];
const int MOD = 1000000007;
int Ord[1005];
inline void Add(int& x, int y) {
  x += y;
  if (x >= MOD) x -= MOD;
}
void precalcC() {
  C[0][0] = 1;
  for (int i = 1; i <= 1000; i++)
    for (int j = 1; j <= i; j++) {
      C[i][j] = C[i - 1][j];
      Add(C[i][j], C[i - 1][j - 1]);
    }
}
void precalcDP() {
  for (int i = 1; i <= M; i++) {
    DP[i][1] = 1;
    Add(Sum[i], 1);
    for (int j = 2; j <= i; j++) {
      DP[i][j] = DP[i - 1][j - 1];
      Add(DP[i][j], (1LL * DP[i - 1][j] * j) % MOD);
      Add(Sum[i], DP[i][j]);
    }
  }
}
inline bool cmp(int a, int b) { return strcmp(S[a] + 1, S[b] + 1) < 0; }
void Read() {
  cin >> M >> N;
  for (int i = 1; i <= N; i++) {
    for (int j = 1; j <= M; j++) cin >> S[j][i];
  }
  for (int i = 1; i <= M; i++) Ord[i] = i;
  sort(Ord + 1, Ord + M + 1, cmp);
}
void Solve() {
  int curr = 1;
  int ans = 1;
  for (int i = 2; i <= M; i++) {
    if (strcmp(S[Ord[i]] + 1, S[Ord[i - 1]] + 1) == 0)
      ++curr;
    else {
      ans = (1LL * ans * Sum[curr]) % MOD;
      curr = 1;
    }
  }
  ans = (1LL * ans * Sum[curr]) % MOD;
  cout << ans << "\n";
}
int main() {
  Read();
  precalcC();
  precalcDP();
  Solve();
  return 0;
}
