#include <bits/stdc++.h>
using namespace std;
template <typename T, typename S>
ostream &operator<<(ostream &os, const pair<T, S> &v) {
  os << "(";
  os << v.first << "," << v.second << ")";
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  os << "[";
  long long sz = v.size();
  for (long long i = 0; i < sz; ++i) {
    os << v[i];
    if (i != sz - 1) os << ",";
  }
  os << "]\n";
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, const set<T> &v) {
  T last = *v.rbegin();
  os << "[";
  for (auto it : v) {
    os << it;
    if (it != last) os << ",";
  }
  os << "]\n";
  return os;
}
template <typename T, typename S>
ostream &operator<<(ostream &os, const map<T, S> &v) {
  for (auto it : v) {
    os << it.first << " : " << it.second << "\n";
  }
  return os;
}
long long power(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b % 2) {
      res = (res * a) % 1000000007;
    }
    b /= 2;
    a = (a * a) % 1000000007;
  }
  return res;
}
template <typename Arg1>
void __f(const char *name, Arg1 &&arg1) {
  cerr << name << " : " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char *names, Arg1 &&arg1, Args &&...args) {
  const char *comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
const long long N = 1e6 + 5;
long long arr[N], Prev[N], temp[N];
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  for (long long i = 1; i <= n; ++i) {
    cin >> arr[i];
    temp[i] = Prev[arr[i]];
    Prev[arr[i]] = i;
  }
  double ans = 1.0 / (double)(n);
  double bla = 0.0;
  for (__typeof(1) i = 1; i <= n; i++) {
    bla += (i - temp[i]) * (n - i + 1);
  }
  bla *= 2;
  bla -= n;
  bla /= (n * n);
  cout << fixed << setprecision(10);
  cout << bla << endl;
  return 0;
}

### Prompt

Your task is to create a cpp solution to the following problem:

You are given an array a consisting of n positive integers. You pick two integer numbers l and r from 1 to n, inclusive (numbers are picked randomly, equiprobably and independently). If l > r, then you swap values of l and r. You have to calculate the expected value of the number of unique elements in segment of the array from index l to index r, inclusive (1-indexed).

Input

The first line contains one integer number n (1 ≤ n ≤ 106). The second line contains n integer numbers a1, a2, ... an (1 ≤ ai ≤ 106) — elements of the array.

Output

Print one number — the expected number of unique elements in chosen segment. 

Your answer will be considered correct if its absolute or relative error doesn't exceed 10 - 4 — formally, the answer is correct if <image>, where x is jury's answer, and y is your answer.

Examples

Input

2
1 2


Output

1.500000


Input

2
2 2


Output

1.000000

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
template <typename T, typename S>
ostream &operator<<(ostream &os, const pair<T, S> &v) {
  os << "(";
  os << v.first << "," << v.second << ")";
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  os << "[";
  long long sz = v.size();
  for (long long i = 0; i < sz; ++i) {
    os << v[i];
    if (i != sz - 1) os << ",";
  }
  os << "]\n";
  return os;
}
template <typename T>
ostream &operator<<(ostream &os, const set<T> &v) {
  T last = *v.rbegin();
  os << "[";
  for (auto it : v) {
    os << it;
    if (it != last) os << ",";
  }
  os << "]\n";
  return os;
}
template <typename T, typename S>
ostream &operator<<(ostream &os, const map<T, S> &v) {
  for (auto it : v) {
    os << it.first << " : " << it.second << "\n";
  }
  return os;
}
long long power(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b % 2) {
      res = (res * a) % 1000000007;
    }
    b /= 2;
    a = (a * a) % 1000000007;
  }
  return res;
}
template <typename Arg1>
void __f(const char *name, Arg1 &&arg1) {
  cerr << name << " : " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char *names, Arg1 &&arg1, Args &&...args) {
  const char *comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
const long long N = 1e6 + 5;
long long arr[N], Prev[N], temp[N];
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  for (long long i = 1; i <= n; ++i) {
    cin >> arr[i];
    temp[i] = Prev[arr[i]];
    Prev[arr[i]] = i;
  }
  double ans = 1.0 / (double)(n);
  double bla = 0.0;
  for (__typeof(1) i = 1; i <= n; i++) {
    bla += (i - temp[i]) * (n - i + 1);
  }
  bla *= 2;
  bla -= n;
  bla /= (n * n);
  cout << fixed << setprecision(10);
  cout << bla << endl;
  return 0;
}

```