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)

#include <stdio.h>
#include <stdlib.h>
  
long long inv;
int *B;
  
long long countup(int *A, int p, int r);
  
int main(int argc, char *argv[])
{
  int n, i, j;
  int *A;
  
  inv = 0;
  
  scanf("%d", &n);
  A = (int *)malloc(sizeof(int) * n);
  B = (int *)malloc(sizeof(int) * n);
  for (i = 0; i < n; i++) scanf("%d", &A[i]);
  
  inv = countup(A, 0, n - 1);
  
  printf("%lld\n", inv);
  
  return 0;
}
  
long long countup(int *A, int p, int r)
{
  int q = p + (r-p)/2;
  int i, j, k, tmp;
  long long ct = 0;
  
  if (r - p < 3) {
    for (i = p; i < r; i++) {
      for (j = r; j > i; j--) {
    if (A[j] < A[j-1]) {
      tmp = A[j-1];
      A[j-1] = A[j];
      A[j] = tmp;
      ct++;
    }
      }
    }
    return ct;
  }
  ct += countup(A, p, q);
  ct += countup(A, q+1, r);
  
  i = p;
  j = q + 1;
  for (k = p; k <= r; k++) {
    if (i >= q + 1 || j >= r + 1) {
      if (i >= q + 1) B[k] = A[j++];
      else B[k] = A[i++];
    } else {
      if (A[i] > A[j]) {
    B[k] = A[j];
    ct += q - i + 1;
    j++;
      } else {
    B[k] = A[i];
    i++;
      }
    }
  }
  for (k = p; k <= r; k++) A[k] = B[k];
  return ct;
}