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;

namespace ALDS1_5_D
{
    class Program
    {
        public long count;

        static void Main(string[] args)
        {
            int N = int.Parse(Console.ReadLine());
            string[] input = Console.ReadLine().Split(' ');
            int[] list = new int[N];
            for (int i = 0; i < N; i++)
            {
                list[i] = int.Parse(input[i]);
            }

            Program prg = new Program();
            prg.mergeSort(list, 0, list.Length);

            Console.WriteLine(prg.count);

        }


        public void merge(int[] list, int left, int middle, int right)
        {
            int n1 = middle - left;
            int n2 = right - middle;
            int[] L = new int[n1 + 1];
            int[] R = new int[n2 + 1];
            for (int i = 0; i < n1; i++)
            {
                L[i] = list[left + i];
            }
            for (int i = 0; i < n2; i++)
            {
                R[i] = list[middle + i];
            }

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

            int p = 0;
            int q = 0;
            for (int i = left; i < right; i++)
            {
                if (L[p] <= R[q])
                {
                    list[i] = L[p];
                    count += q;
                    p++;
                }
                else
                {
                    list[i] = R[q];
                    q++;
                }
            }
        }

        public void mergeSort(int[] list, int left, int right)
        {
            if (left + 1 < right)
            {
                int mid = (left + right) / 2;
                mergeSort(list, left, mid);
                mergeSort(list, mid, right);
                merge(list, left, mid, right);
            }
        }

    }
}