Problem:
MathJax.Hub.Config({ tex2jax: { inlineMath: [["$","$"], ["\\(","\\)"]], processEscapes: true }});

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: Time Limit Exceeded)

#include <stdio.h>
#include <stdlib.h>
#define N 200000 

unsigned int a[N];
int cnt = 0;
void inversion(int* a, int l, int r){
int m = (r+l)/2;
int i = 0, j = 0; 
 if(l == r)
  return;

   inversion(a, l, m);  
   inversion(a, m + 1, r);

    for(i = l; i <= m; i++){
     for(j = m+1; j <= r; j++){
      if(a[i] > a[j]){
        cnt++;
      }   
    }
   }

}



int main(int argc, char** argv){
int n, i = 0;
   scanf("%d", &n);

   for(i; i < n; i++)
    scanf("%d", &a[i]);


   inversion(a, 0, n-1);

   printf("%d\n", cnt); 

  return 0;
}