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>
#pragma GCC optimize(                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                           \
    "-fdelete-null-pointer-checks,inline-functions-called-once,-funsafe-loop-optimizations,-fexpensive-optimizations,-foptimize-sibling-calls,-ftree-switch-conversion,-finline-small-functions,inline-small-functions,-frerun-cse-after-loop,-fhoist-adjacent-loads,-findirect-inlining,-freorder-functions,no-stack-protector,-fpartial-inlining,-fsched-interblock,-fcse-follow-jumps,-fcse-skip-blocks,-falign-functions,-fstrict-overflow,-fstrict-aliasing,-fschedule-insns2,-ftree-tail-merge,inline-functions,-fschedule-insns,-freorder-blocks,-fwhole-program,-funroll-loops,-fthread-jumps,-fcrossjumping,-fcaller-saves,-fdevirtualize,-falign-labels,-falign-loops,-falign-jumps,unroll-loops,-fsched-spec,-ffast-math,Ofast,inline,-fgcse,-fgcse-lm,-fipa-sra,-ftree-pre,-ftree-vrp,-fpeephole2", \
    3)
#pragma G++ optimize("Ofast", 3)
#pragma GCC optimize("Ofast", 3)
#pragma GCC target("sse3", "sse2", "sse")
#pragma GCC target("avx", "sse4", "sse4.1", "sse4.2", "ssse3")
#pragma GCC target("f16c")
#pragma G++ target("sse3", "sse2", "sse")
#pragma G++ target("avx", "sse4", "sse4.1", "sse4.2", "ssse3")
#pragma G++ target("f16c")
inline int read() {
  register int x = 0;
  register char c = getchar();
  register bool m = 0;
  while (!isdigit(c) && c != '-') c = getchar();
  if (c == '-') c = getchar(), m = 1;
  while (isdigit(c)) x = (x << 3) + (x << 1) + c - '0', c = getchar();
  if (m)
    return -x;
  else
    return x;
}
int n, a;
double ans, tmp;
long long c1[100001], c2[100001];
inline void add(int x, int w, long long *c) {
  while (x) {
    c[x] += w;
    x &= x - 1;
  }
}
inline long long query(int x, long long *c) {
  long long ret = 0;
  while (x <= n) {
    ret += c[x];
    x += x & (-x);
  }
  return ret;
}
int main() {
  n = read();
  for (register int i = 1; i <= n; i++) {
    a = read();
    tmp += query(a + 1, c1);
    ans -= (double)query(a + 1, c2) * (n - i + 1);
    ans += (double)(n - i + 1) * i * (i - 1) / 4.0;
    add(a, 1, c1);
    add(a, i, c2);
  }
  printf("%.20lf", (double)(tmp + ans / ((double)n * (n + 1) / 2.0)));
  return 0;
}