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;
using System.Linq;
using System.Text;

namespace ConsoleApplication1
{
    class Program
    {
        static int cnt = 0;

        static void Main(string[] args)
        {
            int n = int.Parse(Console.ReadLine());
            int[] S = Array.ConvertAll(Console.ReadLine().Split(' '), int.Parse);

            MergeSort(S, 0, n);
            
            Console.WriteLine(cnt);
        }

        static void MergeSort(int[] A, int left, int right)
        {
            if (left + 1 < right)
            {
                int mid = (left + right) / 2;
                MergeSort(A, left, mid);
                MergeSort(A, mid, right);
                Merge(A, left, mid, right);
            }
        }

        static void Merge(int[] A,int left,int mid,int right)
        {
            int n1 = mid - left;
            int n2 = right - mid;

            int[] L = new int[n1 + 1];
            int[] R = new int[n2 + 1];

            for (int i = 0; i < n1; i++) L[i] = A[left + i];
            for (int i = 0; i < n2; i++) R[i] = A[mid + i];

            L[n1] = int.MaxValue;
            R[n2] = int.MaxValue;

            int x = 0, y = 0;

            for(int i = left; i < right; i++)
            {
                if (L[x] <= R[y])
                {
                    A[i] = L[x];
                    x++;

                    cnt += y;
                }
                else
                {
                    A[i] = R[y];
                    y++;
                }
            }
        }
    }
}