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;
double eps = 1e-8;
const long double PI = acos((long double)-1);
const long long int INF = 0x3f3f3f3fll;
const int MOD = 1e9 + 7ll;
const int maxn = 1100;
char s[55][maxn];
int inv[maxn], pin_inv[maxn], pin[maxn];
void initCab(int N) {
  inv[0] = pin[0] = pin_inv[0] = inv[1] = pin[1] = pin_inv[1] = 1;
  for (int x = 2; x <= N; x++) {
    inv[x] = (long long int)(MOD - MOD / x) * inv[MOD % x] % MOD;
    pin[x] = (long long int)pin[x - 1] * x % MOD;
    pin_inv[x] = (long long int)pin_inv[x - 1] * inv[x] % MOD;
  }
}
long long int Cab(int n, int m) {
  return (long long int)pin[n] * pin_inv[m] % MOD * pin_inv[n - m] % MOD;
}
long long int bell[1001];
map<long long int, int> pp;
int main() {
  initCab(1001);
  bell[0] = bell[1] = 1;
  for (int x = 2; x <= 1000; x++)
    for (int y = 0; y < x; y++) {
      bell[x] += Cab(x - 1, y) * bell[y];
      bell[x] %= MOD;
    }
  int n, m;
  scanf("%d%d", &m, &n);
  for (int x = 0; x < n; x++) {
    scanf("%s", s[x]);
  }
  for (int x = 0; x < m; x++) {
    long long int t = 0;
    for (int y = 0; y < n; y++) t = (t << 1) + s[y][x] - '0';
    pp[t]++;
  }
  long long int ans = 1;
  for (auto t : pp) {
    ans = ans * bell[t.second] % MOD;
  }
  printf("%I64d\n", ans);
  return 0;
}
