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 MOD = (int)1e9 + 7;
const int FFTMOD = 1007681537;
const int INF = (int)1e9;
const long long LINF = (long long)1e18;
const long double PI = acos((long double)-1);
const long double EPS = 1e-9;
inline long long gcd(long long a, long long b) {
  long long r;
  while (b) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
inline long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
inline long long fpow(long long n, long long k, int p = MOD) {
  long long r = 1;
  for (; k; k >>= 1) {
    if (k & 1) r = r * n % p;
    n = n * n % p;
  }
  return r;
}
template <class T>
inline int chkmin(T& a, const T& val) {
  return val < a ? a = val, 1 : 0;
}
template <class T>
inline int chkmax(T& a, const T& val) {
  return a < val ? a = val, 1 : 0;
}
inline long long isqrt(long long k) {
  long long r = sqrt(k) + 1;
  while (r * r > k) r--;
  return r;
}
inline long long icbrt(long long k) {
  long long r = cbrt(k) + 1;
  while (r * r * r > k) r--;
  return r;
}
inline void addmod(int& a, int val, int p = MOD) {
  if ((a = (a + val)) >= p) a -= p;
}
inline void submod(int& a, int val, int p = MOD) {
  if ((a = (a - val)) < 0) a += p;
}
inline int mult(int a, int b, int p = MOD) { return (long long)a * b % p; }
inline int inv(int a, int p = MOD) { return fpow(a, p - 2, p); }
inline int sign(long double x) { return x < -EPS ? -1 : x > +EPS; }
inline int sign(long double x, long double y) { return sign(x - y); }
void solve() {
  int m, n;
  cin >> m >> n;
  static int f[1005][1005];
  for (int i = (0); i < (n); i++) {
    string s;
    cin >> s;
    for (int j = (0); j < (m); j++)
      for (int k = (j + 1); k < (m); k++)
        if (s[j] != s[k]) {
          f[j][k] = 1;
        }
  }
  static int g[1005];
  static int cc[1005];
  int ptr = 1;
  for (int i = (0); i < (m); i++)
    if (!g[i]) {
      cc[ptr]++;
      g[i] = ptr;
      for (int j = (i + 1); j < (m); j++)
        if (!g[j] && !f[i][j]) {
          cc[ptr]++;
          g[j] = ptr;
        }
      ptr++;
    }
  static int c[1005][1005];
  for (int i = (0); i < (1005); i++) c[0][i] = 1;
  for (int i = (1); i < (1005); i++)
    for (int j = (1); j < (1005); j++)
      c[i][j] = (c[i][j - 1] + c[i - 1][j - 1]) % MOD;
  static int ff[1005];
  ff[0] = ff[1] = 1;
  for (int i = (2); i < (1005); i++) {
    for (int j = (0); j < (i); j++) {
      addmod(ff[i], mult(c[j][i - 1], ff[j]));
    }
  }
  int res = 1;
  for (int i = (1); i < (m + 1); i++) {
    int k = cc[i];
    res = mult(res, ff[k]);
  }
  cout << res << "\n";
}
int main(int argc, char* argv[]) {
  ios_base::sync_with_stdio(0), cin.tie(0);
  if (argc > 1) {
    assert(freopen(argv[1], "r", stdin));
  }
  if (argc > 2) {
    assert(freopen(argv[2], "wb", stdout));
  }
  solve();
  cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
  return 0;
}
