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;
constexpr int mod = 1000000007;
char s[1001];
bitset<1000> t[50];
int cnt[1000];
int comb[1001][1001];
int dp[1001];
bool check(int i, int j, int n) {
  for (int k = 0; k < n; k++)
    if (t[k][i] != t[k][j]) return false;
  return true;
}
int main() {
  int m, n;
  scanf("%d %d", &m, &n);
  for (int i = 0; i < n; i++) {
    scanf("%s", s);
    for (int j = 0; j < m; j++) t[i][j] = s[j] != '0';
  }
  for (int i = 0; i < m; i++) {
    if (cnt[i] == 0) {
      cnt[i] = 1;
      for (int j = i + 1; j < m; j++) {
        if (cnt[j] == 0 && check(i, j, n)) {
          cnt[i]++;
          cnt[j] = -1;
        }
      }
    }
  }
  comb[0][0] = 1;
  for (int i = 1; i <= m; i++) {
    comb[i][0] = 1;
    for (int j = 1; j < i; j++)
      comb[i][j] = (comb[i - 1][j] + comb[i - 1][j - 1]) % mod;
    comb[i][i] = 1;
  }
  dp[0] = 1;
  for (int i = 1; i <= m; i++) {
    for (int k = 1; k <= i; k++)
      dp[i] = (dp[i] + comb[i - 1][k - 1] * 1LL * dp[i - k]) % mod;
  }
  int ans = 1;
  for (int i = 0; i < m; i++) {
    if (cnt[i] > 0) {
      ans = ans * 1LL * dp[cnt[i]] % mod;
    }
  }
  printf("%d\n", ans);
}
