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: Time Limit Exceeded)

#include <stdio.h>
#include <stdlib.h>

long count = 0;

void swap(long *A, long i, long j) {
  long temp;
  temp = A[i];
  A[i] = A[j];
  A[j] = temp;
}

void bubble(long *a, long n) {
  long i,j,*min;
  min = (long*)malloc(sizeof(long) * n);
  min[n-1] = a[n-1];
  for (i = n-2; i > 0; i--) {
    min[i] = (a[i] < min[i+1]) ? a[i] : min[i+1];
  }

  i = 0;  
  while(i < n-1) {
    if(a[i] >= min[i+1]) {
      for (j = n-1; j > i; j--) {
	if(a[j] < a[j-1]) {
	  swap(a,j,j-1);
	  if(a[j] < min[j]) min[j] = a[j];
	  count++;
	}
      }
    }
    i++;
  }
}

int main()
{
  long i,n,*a;
  scanf("%ld", &n);
  a = (long*)malloc(sizeof(long) * n);
  for (i = 0; i < n; i++) {
    scanf("%ld", a+i);
  }
  bubble(a,n);

  printf("%ld\n", count);
  
  return 0;
}