Problem: You play the game with your friend. The description of this game is listed below. 

Your friend creates n distinct strings of the same length m and tells you all the strings. Then he randomly chooses one of them. He chooses strings equiprobably, i.e. the probability of choosing each of the n strings equals <image>. You want to guess which string was chosen by your friend. 

In order to guess what string your friend has chosen, you are allowed to ask him questions. Each question has the following form: «What character stands on position pos in the string you have chosen?» A string is considered guessed when the answers to the given questions uniquely identify the string. After the string is guessed, you stop asking questions. 

You do not have a particular strategy, so as each question you equiprobably ask about a position that hasn't been yet mentioned. Your task is to determine the expected number of questions needed to guess the string chosen by your friend.

Input

The first line contains a single integer n (1 ≤ n ≤ 50) — the number of strings your friend came up with.

The next n lines contain the strings that your friend has created. It is guaranteed that all the strings are distinct and only consist of large and small English letters. Besides, the lengths of all strings are the same and are between 1 to 20 inclusive.

Output

Print the single number — the expected value. Your answer will be considered correct if its absolute or relative error doesn't exceed 10 - 9.

Examples

Input

2
aab
aac


Output

2.000000000000000


Input

3
aaA
aBa
Caa


Output

1.666666666666667


Input

3
aca
vac
wqq


Output

1.000000000000000

Note

In the first sample the strings only differ in the character in the third position. So only the following situations are possible: 

  * you guess the string in one question. The event's probability is <image>; 
  * you guess the string in two questions. The event's probability is <image> · <image> = <image> (as in this case the first question should ask about the position that is other than the third one); 
  * you guess the string in three questions. The event's probability is <image> · <image> · <image> = <image>; 



Thus, the expected value is equal to <image>

In the second sample we need at most two questions as any pair of questions uniquely identifies the string. So the expected number of questions is <image>.

In the third sample whatever position we ask about in the first question, we immediately identify the string.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const string YESNO[2] = {"NO", "YES"};
const string YesNo[2] = {"No", "Yes"};
const string yesno[2] = {"no", "yes"};
void YES(bool t = 1) { cout << YESNO[t] << endl; }
void Yes(bool t = 1) { cout << YesNo[t] << endl; }
void yes(bool t = 1) { cout << yesno[t] << endl; }
template <class T>
using pq = priority_queue<T>;
template <class T>
using pqg = priority_queue<T, vector<T>, greater<T>>;
int scan() { return getchar(); }
void scan(int &a) { cin >> a; }
void scan(long long &a) { cin >> a; }
void scan(char &a) { cin >> a; }
void scan(double &a) { cin >> a; }
void scan(string &a) { cin >> a; }
template <class T, class S>
void scan(pair<T, S> &p) {
  scan(p.first), scan(p.second);
}
template <class T>
void scan(vector<T> &);
template <class T>
void scan(vector<T> &a) {
  for (auto &i : a) scan(i);
}
template <class T>
void scan(T &a) {
  cin >> a;
}
void IN() {}
template <class Head, class... Tail>
void IN(Head &head, Tail &...tail) {
  scan(head);
  IN(tail...);
}
template <class T, class S>
inline bool chmax(T &a, const S &b) {
  return (a < b ? a = b, 1 : 0);
}
template <class T, class S>
inline bool chmin(T &a, const S &b) {
  return (a > b ? a = b, 1 : 0);
}
vector<int> iota(int n) {
  vector<int> a(n);
  iota(begin(a), end(a), 0);
  return a;
}
template <typename T>
vector<int> iota(vector<T> &a, bool greater = false) {
  vector<int> res(a.size());
  iota(begin(res), end(res), 0);
  sort(begin(res), end(res), [&](int i, int j) {
    if (greater) return a[i] > a[j];
    return a[i] < a[j];
  });
  return res;
}
template <class T>
T POW(T x, int n) {
  T res = 1;
  for (; n; n >>= 1, x *= x)
    if (n & 1) res *= x;
  return res;
}
vector<pair<long long, long long>> factor(long long x) {
  vector<pair<long long, long long>> ans;
  for (long long i = 2; i * i <= x; i++)
    if (x % i == 0) {
      ans.push_back({i, 1});
      while ((x /= i) % i == 0) ans.back().second++;
    }
  if (x != 1) ans.push_back({x, 1});
  return ans;
}
template <class T>
vector<T> divisor(T x) {
  vector<T> ans;
  for (T i = 1; i * i <= x; i++)
    if (x % i == 0) {
      ans.push_back(i);
      if (i * i != x) ans.push_back(x / i);
    }
  return ans;
}
template <typename T>
void zip(vector<T> &x) {
  vector<T> y = x;
  sort(begin(y), end(y));
  for (int i = 0; i < x.size(); ++i) {
    x[i] = distance((y).begin(), lower_bound(begin(y), end(y), (x[i])));
  }
}
int popcount(long long x) { return __builtin_popcountll(x); }
int in() {
  int x;
  cin >> x;
  return x;
}
long long lin() {
  unsigned long long x;
  cin >> x;
  return x;
}
template <typename T>
struct edge {
  int from, to;
  T cost;
  int id;
  edge(int to, T cost) : from(-1), to(to), cost(cost) {}
  edge(int from, int to, T cost) : from(from), to(to), cost(cost) {}
  edge(int from, int to, T cost, int id)
      : from(from), to(to), cost(cost), id(id) {}
  edge &operator=(const int &x) {
    to = x;
    return *this;
  }
  operator int() const { return to; }
};
template <typename T>
using Edges = vector<edge<T>>;
using Tree = vector<vector<int>>;
using Graph = vector<vector<int>>;
template <class T>
using Wgraph = vector<vector<edge<T>>>;
Graph getG(int n, int m = -1, bool directed = false, int margin = 1) {
  Tree res(n);
  if (m == -1) m = n - 1;
  while (m--) {
    int a, b;
    cin >> a >> b;
    a -= margin, b -= margin;
    res[a].emplace_back(b);
    if (!directed) res[b].emplace_back(a);
  }
  return move(res);
}
template <class T>
Wgraph<T> getWg(int n, int m = -1, bool directed = false, int margin = 1) {
  Wgraph<T> res(n);
  if (m == -1) m = n - 1;
  while (m--) {
    int a, b;
    T c;
    cin >> a >> b >> c;
    a -= margin, b -= margin;
    res[a].emplace_back(b, c);
    if (!directed) res[b].emplace_back(a, c);
  }
  return move(res);
}
template <class T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  for (auto it = begin(v); it != end(v); ++it) {
    if (it == begin(v))
      os << *it;
    else
      os << " " << *it;
  }
  return os;
}
template <class T, class S>
ostream &operator<<(ostream &os, const pair<T, S> &p) {
  os << p.first << " " << p.second;
  return os;
}
template <class S, class T>
string to_string(pair<S, T> p) {
  return "(" + to_string(p.first) + "," + to_string(p.second) + ")";
}
template <class A>
string to_string(A v) {
  if (v.empty()) return "{}";
  string ret = "{";
  for (auto &x : v) ret += to_string(x) + ",";
  ret.back() = '}';
  return ret;
}
void dump() { cerr << endl; }
template <class Head, class... Tail>
void dump(Head head, Tail... tail) {
  cerr << to_string(head) << " ";
  dump(tail...);
}
template <typename T>
static constexpr T inf = numeric_limits<T>::max() / 2;
struct Setup_io {
  Setup_io() {
    ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    cout << fixed << setprecision(15);
  }
} setup_io;
int main() {
  int n;
  IN(n);
  vector<string> s(n);
  IN(s);
  int m = (int)(s[0]).size();
  long double ans = 0;
  vector<int> g(1 << m);
  vector<long long> f(1 << m);
  for (long long i = 0; i < n; ++i) {
    for (long long j = 0; j < n; ++j) {
      if (j == i) continue;
      int mask = 0;
      for (long long k = 0; k < m; ++k)
        if (s[i][k] == s[j][k]) mask |= 1 << k;
      f[mask] |= 1LL << i;
    }
  }
  for (long long t = 0; t < m; ++t)
    for (long long s = 0; s < 1 << m; ++s)
      if (~s & 1 << t) f[s] |= f[s ^ 1 << t];
  for (long long s = 0; s < 1 << m; ++s) g[s] = popcount(f[s]);
  vector<long double> dp(1 << m);
  dp[0] = 1;
  for (long long mask = 0; mask < 1 << m; ++mask) {
    int c = popcount(mask);
    if (g[mask] == 0) continue;
    for (long long j = 0; j < m; ++j)
      if (~mask & 1 << j) {
        int nxt = mask | 1 << j;
        long double p = (long double)g[nxt] / g[mask];
        ans += (1 - p) * dp[mask] / (m - c) * (c + 1);
        dp[mask | 1 << j] += p * dp[mask] / (m - c);
      }
  }
  cout << ans << '\n';
}