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 INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3f;
const int MOD = (int)1e9 + 7;
const int N = (int)1e3 + 7;
int n, m;
char s[N];
long long col[N], B[N][N], bell[N];
map<long long, int> bit;
void Init() {
  B[0][0] = 1;
  for (int i = (1); i < (m + 1); ++i)
    for (int j = (0); j < (i + 1); ++j)
      B[i][j] = j ? (B[i][j - 1] + B[i - 1][j - 1]) % MOD : B[i - 1][i - 1];
  for (int i = (0); i < (m + 1); ++i) bell[i] = B[i][0];
  for (int i = (0); i < (n); ++i) {
    scanf("%s", s);
    for (int j = (0); j < (m); ++j) col[j] |= (long long)(s[j] == '1') << i;
  }
}
int Solve() {
  for (int j = (0); j < (m); ++j) bit[col[j]]++;
  int ans = 1;
  for (auto o : bit) ans = (ans * bell[o.second]) % MOD;
  return printf("%d\n", ans);
}
int main() {
  while (~scanf("%d%d", &m, &n)) {
    Init();
    Solve();
  }
  return 0;
}
