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 MAXN = 1000;
const int MAXM = 50;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
template <class T>
void Read(T &x) {
  x = 0;
  char c = getchar();
  bool flag = 0;
  while (c < '0' || '9' < c) {
    if (c == '-') flag = 1;
    c = getchar();
  }
  while ('0' <= c && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  if (flag) x = -x;
}
map<long long int, int> mp;
long long int s[MAXN + 10];
char str[MAXN + 10];
int n, m, cnt;
int num[MAXN + 10];
int dp[MAXN + 10][MAXN + 10];
int dfs(int len, int block) {
  if (dp[len][block] != -1) return dp[len][block];
  if (block == 1)
    return 1;
  else {
    dp[len][block] = 0;
    if (len > block)
      dp[len][block] =
          (dp[len][block] + 1ll * dfs(len - 1, block) * block % MOD) % MOD;
    dp[len][block] = (dp[len][block] + dfs(len - 1, block - 1)) % MOD;
    return dp[len][block];
  }
}
int main() {
  Read(n), Read(m);
  for (int i = 1; i <= m; ++i) {
    scanf("%s", str + 1);
    for (int j = 1; j <= n; ++j)
      if (str[j] == '1') s[j] |= 1ll << (i - 1);
  }
  memset(num, 0, sizeof(num));
  mp.clear(), cnt = 0;
  for (int i = 1; i <= n; ++i) {
    if (!mp[s[i]]) mp[s[i]] = ++cnt;
    ++num[mp[s[i]]];
  }
  memset(dp, -1, sizeof(dp));
  int ans = 1, t;
  for (int i = 1; i <= cnt; ++i) {
    t = 0;
    for (int j = 1; j <= num[i]; ++j) t = (t + dfs(num[i], j)) % MOD;
    ans = 1ll * ans * t % MOD;
  }
  printf("%d\n", ans);
}
