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: Accepted)

using System;
using System.Collections;
					
public class Program
{
	public static long comp = 0;
	public static void Main()
	{
		int n = int.Parse(Console.ReadLine());
		int[] ls = Array.ConvertAll(Console.ReadLine().Trim().Split(' '), int.Parse);
		mergeSort(ls, 0, n);
		Console.WriteLine(comp);
	}
	public static void mergeSort(int[] ls, int left, int right){
		if (left + 1 < right){
			int mid = (left + right)/2;
			mergeSort(ls, left, mid);
			mergeSort(ls, mid, right);
			merge(ls, left, mid, right);
		}
	} 
	public static void merge(int[] ls, int left, int mid, int right){
		int n1 = mid - left, n2 = right - mid;
		int[] L = new int[n1 + 1], R = new int[n2 + 1];
		for(int i = 0; i < n1; i++) L[i] = ls[left + i];
		for(int i = 0; i < n2; i++) R[i] = ls[mid + i];
		L[n1] = R[n2] = int.MaxValue;
		for(int k = left, i = 0, j = 0; k < right; k++) 
			if (L[i] < R[j]){
				ls[k] = L[i++];
			}
			else{
				ls[k] = R[j++];
				comp += n1 - i;
			}
	}
}