Problem: You are given a permutation of integers from 1 to n. Exactly once you apply the following operation to this permutation: pick a random segment and shuffle its elements. Formally:

  1. Pick a random segment (continuous subsequence) from l to r. All <image> segments are equiprobable. 
  2. Let k = r - l + 1, i.e. the length of the chosen segment. Pick a random permutation of integers from 1 to k, p1, p2, ..., pk. All k! permutation are equiprobable. 
  3. This permutation is applied to elements of the chosen segment, i.e. permutation a1, a2, ..., al - 1, al, al + 1, ..., ar - 1, ar, ar + 1, ..., an is transformed to a1, a2, ..., al - 1, al - 1 + p1, al - 1 + p2, ..., al - 1 + pk - 1, al - 1 + pk, ar + 1, ..., an. 



Inversion if a pair of elements (not necessary neighbouring) with the wrong relative order. In other words, the number of inversion is equal to the number of pairs (i, j) such that i < j and ai > aj. Find the expected number of inversions after we apply exactly one operation mentioned above.

Input

The first line contains a single integer n (1 ≤ n ≤ 100 000) — the length of the permutation.

The second line contains n distinct integers from 1 to n — elements of the permutation.

Output

Print one real value — the expected number of inversions. Your answer will be considered correct if its absolute or relative error does not exceed 10 - 9. 

Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.

Example

Input

3
2 3 1


Output

1.916666666666666666666666666667

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 10;
const long long INF = 0x3f3f3f3f3f3f3f3f;
long long a[N], b[N], c[N];
int n;
int main() {
  auto lowbit = [&](int x) { return x & -x; };
  auto update = [&](int x, int val) {
    for (; x <= n; x += lowbit(x)) c[x]++, b[x] += val;
  };
  auto get_sum = [&](int x) {
    long long ans1 = 0, ans2 = 0;
    for (; x; x -= lowbit(x)) ans1 += c[x], ans2 += b[x];
    return make_pair(ans1, ans2);
  };
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
  double ans = 0;
  for (int i = 1; i < n; i++) {
    long long sum = (n - i + 1) * 1ll * (n - i) / 2;
    ans += sum * i / (n * 1ll * (n + 1) + 0.0);
  }
  for (int i = 1; i <= n; i++) {
    pair<long long, long long> res1 = get_sum(a[i]);
    pair<long long, long long> res2 = get_sum(n);
    long long sum1 = res2.second - res1.second;
    sum1 = sum1 * (n - i + 1);
    long long sum2 = (res2.first - res1.first) * 1ll * n * (n + 1) / 2;
    ans += (sum2 - sum1) / (n * 1ll * (n + 1) / 2 + 0.0);
    update(a[i], i);
  }
  printf("%.10f\n", ans);
  return 0;
}