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.Generic;
using System.Text;

class Program
{
    public void Proc()
    {

        int[] A = new int[MAX];
        int n = int.Parse(Console.ReadLine());
        string[] inpt = Console.ReadLine().Split(' ');
        for(int i=0; i<n; i++) {
            A[i] = int.Parse(inpt[i]);
        }
        long ans = this.MergeSort(A, n, 0, n);
        Console.WriteLine(ans);
    }

    private long Merge(int[] A, int n, int left, int mid, int right)
    {
        long cnt = 0;
        int n1 = mid - left;
        int n2 = right - mid;
        for (int i = 0; i < n1; i++)
        {
            this.L[i] = A[left + i]; 
        }
        for (int i = 0; i < n2; i++)
        {
            this.R[i] = A[mid + i];
        }
        this.L[n1] = SENTINEL;
        this.R[n2] = SENTINEL;
        int j = 0;
        int k = 0;
        for (int i = left; i < right; i++)
        {
            if (this.L[j] <= this.R[k])
            {
                A[i] = this.L[j++];
            }
            else
            {
                A[i] = this.R[k++];
                cnt += (n1 - j);
            }
        }
        return cnt;
    }

    private long MergeSort(int[] A, int n, int left, int right) {
        int mid;
        long v1, v2, v3;
        if (left + 1 < right)
        {
            mid = (left + right) / 2;
            v1 = this.MergeSort(A, n, left, mid);
            v2 = this.MergeSort(A, n, mid, right);
            v3 = this.Merge(A, n, left, mid, right);
            return v1 + v2 + v3;
        }
        else
        {
            return 0;
        }
    }

    private const int MAX = 200000;
    private const int SENTINEL = int.MaxValue;
    private int[] L = new int[MAX / 2 + 2];
    private int[] R = new int[MAX / 2 + 2];


    static void Main(string[] args)

    {

        Program prg = new Program();
        prg.Proc();

    }
}