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;
struct node {
  long long x, y, cnt, num, sum;
  node *L, *R;
  node(long long X, long long Num) {
    x = X;
    y = rand() * rand();
    cnt = 1;
    num = sum = Num;
    L = R = nullptr;
  }
};
long long cnt(node *T) {
  if (T) return T->cnt;
  return 0;
}
long long sum(node *T) {
  if (T) return T->sum;
  return 0;
}
void update(node *T) {
  if (T) {
    T->sum = sum(T->L) + T->num + sum(T->R);
    T->cnt = cnt(T->L) + 1 + cnt(T->R);
  }
}
node *nmerge(node *L, node *R) {
  if (!L) return R;
  if (!R) return L;
  if (L->y > R->y) {
    L->R = nmerge(L->R, R);
    update(L);
    update(R);
    return L;
  } else {
    R->L = nmerge(L, R->L);
    update(L);
    update(R);
    return R;
  }
}
pair<node *, node *> split(node *T, long long x) {
  if (!T) return {nullptr, nullptr};
  if (T->x <= x) {
    auto tt = split(T->R, x);
    T->R = tt.first;
    update(T);
    return {T, tt.second};
  } else {
    auto tt = split(T->L, x);
    T->L = tt.second;
    update(T);
    return {tt.first, T};
  }
}
long long n;
long double den, dif = 0, last = 0, invNow = 0;
vector<long long> v;
node *T = nullptr;
long double countInversion() {
  long long ret = 0;
  node *TI = nullptr;
  for (long long i = 0; i < n; i++) {
    auto tt = split(TI, v[i]);
    ret += cnt(tt.second);
    TI = nmerge(tt.first, new node(v[i], 1));
    TI = nmerge(TI, tt.second);
  }
  return (long double)ret;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(nullptr);
  cout.precision(16);
  cin >> n;
  den = n * (n + 1) / 2;
  v.resize(n);
  for (long long i = 0; i < n; i++) cin >> v[i];
  for (long long i = 0; i < n; i++) {
    last += (long double)i * (i + 1) / 4;
    dif += last;
    auto tt = split(T, v[i]);
    invNow += sum(tt.second);
    T = nmerge(tt.first, new node(v[i], i + 1));
    T = nmerge(T, tt.second);
    dif -= invNow;
  }
  cout << countInversion() + dif / den;
  return 0;
}