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;
bool debug = 0;
int n, m, k;
int dx[4] = {0, 1, 0, -1}, dy[4] = {1, 0, -1, 0};
string direc = "RDLU";
long long ln, lk, lm;
void etp(bool f = 0) {
  puts(f ? "YES" : "NO");
  exit(0);
}
void addmod(int &x, int y, int mod = 1000000007) {
  assert(y >= 0);
  x += y;
  if (x >= mod) x -= mod;
  assert(x >= 0 && x < mod);
}
void et() {
  puts("-1");
  exit(0);
}
long long fastPow(long long x, long long y, int mod = 1000000007) {
  long long ans = 1;
  while (y > 0) {
    if (y & 1) ans = (x * ans) % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return ans;
}
long long gcd1(long long x, long long y) {
  long long z = y;
  while (x % y != 0) {
    z = x % y;
    x = y;
    y = z;
  }
  return z;
}
map<long long, int> mp;
char s[55][1005];
int S[1005][1005];
void init(int n, int mod) {
  S[0][0] = 1;
  for (int(i) = 1; (i) <= (int)(n); (i)++) {
    for (int j = 1; j <= i; j++) {
      S[i][j] = (S[i - 1][j - 1] + (long long)S[i - 1][j] * (j) % mod) % mod;
    }
  }
}
void fmain(int ID) {
  scanf("%d%d", &m, &n);
  init(m, 1000000007);
  for (int(i) = 1; (i) <= (int)(n); (i)++) scanf("%s", s[i] + 1);
  for (int(j) = 1; (j) <= (int)(m); (j)++) {
    long long c = 0;
    for (int(i) = 1; (i) <= (int)(n); (i)++)
      c |= ((long long)(s[i][j] - '0')) << i;
    mp[c]++;
  }
  int ans = 1;
  for (auto p : mp) {
    int a = p.second;
    int cc = 0;
    for (int(i) = 1; (i) <= (int)(a); (i)++) addmod(cc, S[a][i]);
    ans = (long long)ans * cc % 1000000007;
  }
  printf("%d\n", ans);
}
int main() {
  int t = 1;
  for (int(i) = 1; (i) <= (int)(t); (i)++) {
    fmain(i);
  }
  return 0;
}
