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;
inline long long ri() {
  long long x;
  scanf("%lld", &x);
  return x;
}
template <typename T>
inline bool smax(T& a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <typename T>
inline bool smin(T& a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
struct mint {
  const static long long mod = 1e9 + 7;
  int x;
  mint() : x(0) {}
  mint(long long x) : x(((x % mod) + mod) % mod){};
  mint operator+=(const mint& a) {
    if ((x += a.x) >= mod) x -= mod;
    return *this;
  }
  mint operator-=(const mint& a) {
    if ((x += mod - a.x) >= mod) x -= mod;
    return *this;
  }
  mint operator*=(const mint& a) {
    x = ((long long)x * a.x) % mod;
    return *this;
  }
  mint operator*(const mint& a) { return mint(*this) *= a; }
  mint operator-(const mint& a) { return mint(*this) -= a; }
  mint operator+(const mint& a) { return mint(*this) += a; }
  template <typename T>
  mint operator*(const T& a) {
    return mint(*this) *= a;
  }
  template <typename T>
  mint operator-(const T& a) {
    return mint(*this) -= a;
  }
  template <typename T>
  mint operator+(const T& a) {
    return mint(*this) += a;
  }
  template <typename T>
  friend mint operator*(T x, const mint& m) {
    return mint(x) *= m;
  }
  template <typename T>
  friend mint operator-(T x, const mint& m) {
    return mint(x) -= m;
  }
  template <typename T>
  friend mint operator+(T x, const mint& m) {
    return mint(x) += m;
  }
  bool operator==(const mint& a) const { return x == a.x; }
  operator long long() const { return x; }
};
mint pw(mint a, long long b) {
  mint res(1);
  while (b) {
    if (b & 1) res *= a;
    a *= a;
    b >>= 1;
  }
  return res;
}
mint inv(mint a) { return pw(a, mint::mod - 2); }
const int maxn = 1200;
mint p[maxn], q[maxn], c[maxn][maxn];
set<string> sv;
int n, m;
int main() {
  for (int i = 0; i < (maxn); i++) {
    c[i][0] = c[i][i] = 1;
    for (int j = (1); j < (i); j++) c[i][j] = c[i - 1][j] + c[i - 1][j - 1];
  }
  q[0] = 1;
  for (int i = (1); i < (maxn); i++) {
    q[i] = q[i - 1];
    for (int j = (1); j < (i + 1); j++) {
      q[i] += q[i - j] * c[i - 1][j - 1];
    }
  }
  p[0] = 1;
  for (int i = (1); i < (maxn); i++) {
    for (int j = (1); j < (i + 1); j++) {
      p[i] += p[i - j] * c[i - 1][j - 1];
    }
  }
  cin >> m >> n;
  for (int i = 0; i < (n); i++) {
    string s;
    cin >> s;
    sv.insert(s);
  }
  sv.insert(string(m, '1'));
  map<string, int> mp;
  vector<string> v(sv.begin(), sv.end());
  for (int k = (0); k < (m); k++) {
    string key;
    for (int i = 0; i < (v.size()); i++) {
      key += v[i][k];
    }
    mp[key]++;
  }
  mint res = 1;
  for (auto& it : mp) {
    res *= p[it.second];
  }
  cout << res << endl;
  return 0;
}
