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: Time Limit Exceeded)

using System;

namespace ALDS1_5_D
{
    class Program
    {
        static long count = 0;

        static void Main(string[] args)
        {
            int n = int.Parse(Console.ReadLine());
            int[] array = Array.ConvertAll(Console.ReadLine().Split(' '), int.Parse);
            Comp(array, n - 1, n - 1, 0);
            Console.WriteLine(count);
        }

        static void Comp(int[] array, int x, int y, int cnt)
        {
            if (array[x - 1] > array[x])
            {
                count++;
                int tmp = array[x];
                array[x] = array[x - 1];
                array[x - 1] = tmp;
                
            }
            if (x == 1) return;
            if (y == cnt || y == 1) return;

            Comp(array, x - 1, y, cnt + 1);
            Comp(array, x, y - 1, 0);
            
        }
    }
}