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 JAVA (judge verdict: Time Limit Exceeded)

import java.util.*;
import java.io.*;
class Main{
  static int count =0;
  static int n=0;
  public static void main(String args[])throws Exception{
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    n = Integer.parseInt(br.readLine());
    String str [] = br.readLine().split("\\s+");
    int array [] = new int[n];
    for(int i = 0; i<n; i++){
      array[i] = Integer.parseInt(str[i]);
    }
    Inversions(array);
    System.out.println(count);
  }
  //TLE
   static void Inversions(int A[]){
     for(int i = 0; i<n-1; i++){
       for(int j = i+1;j<n;j++){
           if(A[i]>A[j])count++;
       }
     }
   }
}