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;
using System.Linq;

namespace ALDS1_5_D {
    class Problem {
        private int[] _nums, _workL, _workR;
        private long _count;

        public Problem(int[] nums,int n) {
            _nums = nums;

            int capacity = n / 2 + 2;
            _workL = new int[capacity];
            _workR = new int[capacity];
            
            _count = 0;
            MergeSort(0, n);
        }

        public long Count {
            get { return _count; }
        }
        
        private void MergeSort(int left,int right) {
            if (left + 1 < right) {
                int mid = (left + right) / 2;
                MergeSort(left, mid);
                MergeSort(mid, right);
                Merge(left, mid, right);
            }
        }

        private void Merge(int left,int mid,int right) {
            int lenL = mid - left;
            for(int i = 0; i < lenL; i++)
                _workL[i] = _nums[left + i];
            _workL[lenL] = int.MaxValue;

            int lenR = right - mid;
            for(int i = 0; i < lenR; i++)
                _workR[i] = _nums[mid + i];
            _workR[lenR] = int.MaxValue;

            int indexL = 0;
            int indexR = 0;
            for(int i = left; i < right; i++) {
                if (_workL[indexL] <= _workR[indexR]) {
                    _nums[i] = _workL[indexL];
                    indexL++;
                    _count += indexR;
                }
                else {
                    _nums[i] = _workR[indexR];
                    indexR++;
                }
            }
        }
    }

    class Program {
        static void Main(string[] args) {
            int n = int.Parse(Console.ReadLine());
            int[] arr = Console.ReadLine().Split(' ').Select(int.Parse).ToArray();
            Problem p = new Problem(arr, n);
            Console.WriteLine(p.Count);
        }
    }
}