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;
using System.Collections.Generic;
using System.Linq;
using System.Text;

namespace aizu1
{
    public class Solution
    {
        public static void Main(string[] args)
        {
             Console.ReadLine();
            int[] s = Array.ConvertAll<string, int>(Console.ReadLine().Split(' '), int.Parse);
            long comCount = mergeSort(s, 0, s.Length);
            Console.WriteLine(comCount);
        }


        public static long mergeSort(int[] list, int left, int right)
        {
            if (left == right - 1) return 0;

            return mergeSort(list, left, (right + left) / 2) + mergeSort(list, (right + left) / 2, right) + merge(list, left, (left + right) / 2, right);
        }

        public static long merge(int[] list, int left, int mid, int right)
        {
            int[] L = new int[mid - left + 1];
            int[] R = new int[right - mid + 1];
            int com = 0;

            for (int m = 0; m < mid - left; m++)
            {
                L[m] = list[m + left];
            }

            for (int i = 0; i < right - mid; i++)
            {
                R[i] = list[mid + i];
            }

            L[mid - left] = int.MaxValue;
            R[right - mid] = int.MaxValue;

            int j = 0, k = 0;
            for (int l = left; l < right; l++)
            {
                if (L[j] <= R[k])
                    list[l] = L[j++];
                else
                {
                    list[l] = R[k++];
                    com += mid - left - j;
                }
            }
            return com;
        }
    }
}