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;//リストの使用
using System.Collections.Generic;
class Program
{
static int n = int.Parse(Console.ReadLine());
static long[] nums = Array.ConvertAll(Console.ReadLine().Split(' '),long.Parse);
static long[] bits = new long[n+1];

	static void Main()
	{
    long[][] vertexes = new long[n][];//宣言
    for(int i = 0; i < n; i++) vertexes[i] = new long[]{i,nums[i]};//代入
    Array.Sort(vertexes, (a, b) => a[1].CompareTo(b[1]));
    for(int i = 0; i < n; i++) vertexes[i][1] = i+1;
    Array.Sort(vertexes, (a, b) => a[0].CompareTo(b[0]));
    for(int i = 0; i < n; i++) nums[i] = vertexes[i][1];
  
    long answer = 0;
    for(long i = 0; i < n; i++)
    {
      answer += i - BitSum(nums[i]);
      BitAdd(nums[i], 1);
    }
    
		Console.WriteLine(answer);
	}

  static void BitAdd(long i, long x)//BITでiだけにx加算
  {
    while(i <= n)
    {
      bits[i] += x;
      i += i&-i;
    }
  }

  static long BitSum(long i)//BITでi以下の総和を計算
  {
    long sum = 0;
    while(i > 0)
    {
      sum += bits[i];
      i -= i&-i;
    }
    return sum;
  }
  
}