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: Wrong Answer)

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

unsigned int a[N];

int L[N/2 + 2], R[N/2 + 2];

int merge_sort(int *a, int l, int r){
int m  = (l+r)/2, count, i, j;
int n1, n2, v1, v2;
long long cnt = 0;
  if(l == r)
   return 0;

  v1 = merge_sort(a, l, m);
  v2 = merge_sort(a, m + 1, r);
  n1 = m-l+1;
  n2 = r - m;
  L[n1]  = R[n2] = 200000000;
  for(count = l; count <= m; count++){
     L[count - l] = a[count];
  }
  for(count = m + 1; count <= r; count++){
    R[count - m - 1] = a[count];
  }
  for(i = 0, j = 0, count = l; count <= r; count++){
    if(L[i] <= R[j]){
     a[count] = L[i++];
    }
    else{
     a[count] = R[j++];
     cnt += n1 - i;
    }

  }

   return cnt + v1 + v2;
}

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

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


 printf("%d\n", merge_sort(a, 0, n-1));

  return 0;
}