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 msk[55][1000 + 5];
int dp[1000 + 5], C[1000 + 5][1000 + 5];
int main() {
  for (int i = 0; i <= 1000; i++) C[i][0] = 1;
  for (int i = 1; i <= 1000; i++)
    for (int j = 1; j <= i; j++)
      C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % 1000000007;
  int m, n;
  scanf("%d%d", &m, &n);
  for (int i = 0; i <= n - 1; i++) scanf("%s", msk[i] + 1);
  dp[0] = 1;
  for (int i = 1; i <= m; i++)
    for (int j = 0; j <= i - 1; j++)
      dp[i] = (dp[i] + 1ll * C[i - 1][j] * dp[i - 1 - j]) % 1000000007;
  map<long long, int> cnt;
  for (int i = 1; i <= m; i++) {
    long long x = 0;
    for (int j = 0; j <= n - 1; j++)
      if (msk[j][i] == '1') x |= 1ll << j;
    cnt[x]++;
  }
  long long ans = 1;
  for (auto [x, c] : cnt) ans = ans * dp[c] % 1000000007;
  printf("%lld\n", ans);
  return 0;
}
