Problem:
MathJax.Hub.Config({ tex2jax: { inlineMath: [["$","$"], ["\\(","\\)"]], processEscapes: true }});

The Number of Inversions

For a given sequence $A = \{a_0, a_1, ... a_{n-1}\}$, the number of pairs $(i, j)$ where $a_i > a_j$ and $i < j$, is called the number of inversions. The number of inversions is equal to the number of swaps of Bubble Sort defined in the following program:

bubbleSort(A)
cnt = 0 // the number of inversions
for i = 0 to A.length-1
for j = A.length-1 downto i+1
if A[j] < A[j-1]
swap(A[j], A[j-1])
cnt++

return cnt

For the given sequence $A$, print the number of inversions of $A$. Note that you should not use the above program, which brings Time Limit Exceeded.

Input

In the first line, an integer $n$, the number of elements in $A$, is given. In the second line, the elements $a_i$ ($i = 0, 1, .. n-1$) are given separated by space characters.

output

Print the number of inversions in a line.

Constraints

$ 1 \leq n \leq 200,000$

$ 0 \leq a_i \leq 10^9$

$a_i$ are all different

Sample Input 1

5
3 5 2 1 4

Sample Output 1

6

Sample Input 2

3
3 1 2

Sample Output 2

2

Solution in Ruby (judge verdict: Time Limit Exceeded)

#!/usr/bin/ruby
# -*- coding: utf-8 -*-
=begin
# ALDS1_5-D

=end

def cnt(l)
  n = l.length
  q = 0
# ???????????????????????????????????????????????????????????????????????£???????????°?????????
  0.upto(l.length-1) {|i| 
      (i+1).upto(n-1) {|j|
        if l[j] < l[i] then
          q += 1
        end
      }
  }
  return q
end

def mkidx(f)
  g = []
  0.upto(f.length-1){|i| g.push([f[i], i])}
  h = g.sort

  e=[]
  0.upto(h.length-1){|i| e[h[i][1]]=i}

  return e
end

class Main
  def initialize()
    #fh = open("/home/muz/Downloads/ALDS1_5_B_in0.txt")
    #fh = open("/home/muz/Downloads/ALDS1_5_D_in3.txt")
    fh = STDIN
    n = fh.gets.to_i
    a = fh.gets.chomp.split.map{|x| x.to_i}
    fh.close
    idx = mkidx(a)
    n = cnt(idx)

    puts n
  end
end

if $0 == __FILE__ then
  Main.new
end