Problem:
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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 C# (judge verdict: Wrong Answer)

using System;
using System.Collections.Generic;

namespace Project
{
	class MainClass
	{
		public static void Main(string[] args)
		{
			int n = int.Parse(Console.ReadLine());
			int[] a = Array.ConvertAll(Console.ReadLine().Split(' '), int.Parse);
			int count = 0;
			int min = a[n -1];
			int max = a[n - 1];
			for (int i = n - 1; i >= 0; i--)
			{
				if (a[i] > max)
				{
					max = a[i];
					count += n - 1 - i;
					continue;
				}
				else if (a[i] < min)
				{
					min = a[i];
					continue;
				}
				for (int j = i + 1; j < n; j++)
				{
					if (a[i] > a[j])
						count++;
				}
			}
			Console.WriteLine(count);
		}
	}
}