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;
int n, m;
long long mod = 1e9 + 7;
char a[51][1001];
long long dp[1001][1001], num[1001];
int main() {
  long long x, ans = 1;
  scanf("%d%d", &m, &n);
  for (int i = 0; i < n; i++) scanf("%s", &a[i]);
  map<long long, long long> mp;
  dp[0][0] = 1;
  for (int i = 1; i <= 1000; i++)
    for (int j = 1; j <= 1000; j++) {
      dp[i][j] = (dp[i - 1][j] * j + dp[i - 1][j - 1]) % mod;
      num[i] = (num[i] + dp[i][j]) % mod;
    }
  for (int i = 0; i < m; i++) {
    x = 0;
    for (int j = 0; j < n; j++) x = x * 2 + (a[j][i] - '0');
    mp[x]++;
  }
  for (auto pp : mp) ans = ans * num[pp.second] % mod;
  printf("%d\n", ans);
}
