#include <bits/stdc++.h>
using namespace std;
void pass() { return; }
const double EPS = 1e-6;
mt19937 rnd(rand() << 15 + rand());
const long long MOD = 1e9 + 7;
const long long BIG_MOD = 1e18 + 7;
const long long MAXN = 1e6 + 7;
long long a[MAXN];
long long last[MAXN];
long long myright[MAXN];
signed main() {
  cout.tie(NULL);
  cout.precision(20);
  ios_base::sync_with_stdio(false);
  long long n;
  cin >> n;
  if (n == 1) {
    cout << "1\n";
    return 0;
  }
  for (long long i = 0; i < MAXN; ++i) {
    last[i] = n;
  }
  for (long long i = 0; i < n; ++i) {
    cin >> a[i];
  }
  for (long long i = n - 1; i >= 0; --i) {
    myright[i] = last[a[i]] - i - 1;
    last[a[i]] = i;
  }
  long long ans = 0;
  for (long long i = 0; i < n; ++i) {
    ans += (i + 1) * (myright[i] + 1) - 1;
  }
  double sr = (double)ans / (n * (n - 1) / 2);
  cout << (double)(sr * n * (n - 1) + n) / (n * n);
  return 0;
}

### Prompt

Create a solution in cpp for 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;
void pass() { return; }
const double EPS = 1e-6;
mt19937 rnd(rand() << 15 + rand());
const long long MOD = 1e9 + 7;
const long long BIG_MOD = 1e18 + 7;
const long long MAXN = 1e6 + 7;
long long a[MAXN];
long long last[MAXN];
long long myright[MAXN];
signed main() {
  cout.tie(NULL);
  cout.precision(20);
  ios_base::sync_with_stdio(false);
  long long n;
  cin >> n;
  if (n == 1) {
    cout << "1\n";
    return 0;
  }
  for (long long i = 0; i < MAXN; ++i) {
    last[i] = n;
  }
  for (long long i = 0; i < n; ++i) {
    cin >> a[i];
  }
  for (long long i = n - 1; i >= 0; --i) {
    myright[i] = last[a[i]] - i - 1;
    last[a[i]] = i;
  }
  long long ans = 0;
  for (long long i = 0; i < n; ++i) {
    ans += (i + 1) * (myright[i] + 1) - 1;
  }
  double sr = (double)ans / (n * (n - 1) / 2);
  cout << (double)(sr * n * (n - 1) + n) / (n * n);
  return 0;
}

```