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.Linq;

namespace Recursion___Divide_and_Conquer
{
    class The_Number_of_Inversions
    {
        static void Main(string[] args)
        {
            Console.ReadLine();
            var A = Console.ReadLine().Split(' ').Select(s => int.Parse(s)).ToArray();
            var c = MergeSort(A, 0, A.Length);
            Console.WriteLine(c);
        }

        static int Merge(int[] A, int left, int mid, int right)
        {
            var c = 0;
            var n1 = mid - left;
            var n2 = right - mid;
            var L = new int[n1 + 1];
            var 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] = R[n2] = int.MaxValue;
            var p = 0; var q = 0;
            for (int k = left; k < right; k++)
                if (L[p] <= R[q]) A[k] = L[p++];
                else { c += L.Length - 1 - p; A[k] = R[q++]; }
            return c;
        }

        static int MergeSort(int[] A, int left, int right)
        {
            var c = 0;
            if (left + 1 < right)
            {
                var mid = (left + right) / 2;
                c += MergeSort(A, left, mid);
                c += MergeSort(A, mid, right);
                c += Merge(A, left, mid, right);
            }
            return c;
        }
    }
}