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.Collections;
using System.Collections.Generic;
using System.Diagnostics;
using System.IO;
using System.Linq;
using System.Numerics;
using System.Text;
using System.Text.RegularExpressions;
using System.Threading.Tasks;
using static System.Math;
using MethodImplAttribute = System.Runtime.CompilerServices.MethodImplAttribute;
using MethodImplOptions = System.Runtime.CompilerServices.MethodImplOptions;

public static class P
{
    public static void Main()
    {
        int n = int.Parse(Console.ReadLine());
        var a = Console.ReadLine().Split().Select(int.Parse).ToArray();
        var dic = a.OrderBy(x => x).Distinct().Select((elem, ind) => new { elem, ind }).ToDictionary(x => x.elem, x => x.ind);
        a = a.Select(x => dic[x]).ToArray();
        SegmentTree<int> segTree = new SegmentTree<int>(dic.Count, 0, (x, y) => x + y);
        long res = 0;
        for (int i = 0; i < a.Length; i++)
        {
            res += segTree.Query(a[i] + 1, segTree.Size - 1);
            segTree[a[i]]++;
        }
        Console.WriteLine(res);
    }
}


class SegmentTree<T>
{
    public int Size { get; private set; }
    T Identity;
    T[] Data;
    Func<T, T, T> Merge;
    int LeafCount;

    [MethodImpl(MethodImplOptions.AggressiveInlining)]
    public SegmentTree(int size, T identity, Func<T, T, T> merge)
    {
        Init(size, identity, merge);
        for (int i = 1; i < Data.Length; i++) Data[i] = identity;
    }
    [MethodImpl(MethodImplOptions.AggressiveInlining)]
    public SegmentTree(T[] elems, T identity, Func<T, T, T> merge)
    {
        Init(elems.Length, identity, merge);
        elems.CopyTo(Data, LeafCount);
        for (int i = elems.Length + LeafCount; i < Data.Length; i++) Data[i] = identity;
        for (int i = LeafCount - 1; i >= 1; i--) Data[i] = Merge(Data[i << 1], Data[(i << 1) | 1]);
    }
    [MethodImpl(MethodImplOptions.AggressiveInlining)]
    private void Init(int size, T identity, Func<T, T, T> merge)
    {
        Size = size;
        Identity = identity;
        Merge = merge;
        LeafCount = 1;
        while (LeafCount < size) LeafCount <<= 1;
        Data = new T[LeafCount << 1];
    }

    public T this[int index]
    {
        [MethodImpl(MethodImplOptions.AggressiveInlining)]
        get { return Data[LeafCount + index]; }
        [MethodImpl(MethodImplOptions.AggressiveInlining)]
        set { Update(index, value); }
    }
    [MethodImpl(MethodImplOptions.AggressiveInlining)]
    public void Update(int i, T x) { Data[i += LeafCount] = x; while (0 < (i >>= 1)) Data[i] = Merge(Data[i << 1], Data[(i << 1) | 1]); }
    [MethodImpl(MethodImplOptions.AggressiveInlining)]
    public void Operate(int i, T x) { Data[i += LeafCount] = Merge(Data[i], x); while (0 < (i >>= 1)) Data[i] = Merge(Data[i << 1], Data[(i << 1) | 1]); }
    [MethodImpl(MethodImplOptions.AggressiveInlining)]
    public T Query(int l, int r)
    {
        T lRes = Identity, rRes = Identity;
        for (l += LeafCount, r += LeafCount; l <= r; l = (l + 1) >> 1, r = (r - 1) >> 1)
        {
            if ((l & 1) == 1) lRes = Merge(lRes, Data[l]); if ((r & 1) == 0) rRes = Merge(Data[r], rRes);
        }
        return Merge(lRes, rRes);
    }
}