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 <limits.h>

#define MAX 200000
#define INF INT_MAX
typedef long long ll;

void merge(int*, int, int, int);
void mergesort(int*, int, int);

ll ans;

int n, a[MAX];

int main(){
  int i;

  scanf("%d" ,&n);
  for(i = 0 ; i < n ; i++){
    scanf("%d" ,a + i);
  }
  mergesort(a,0,n);
  printf("%lld\n" ,ans);

  return 0;
}

void merge(int a[], int left, int mid, int right){
  int n1 = mid - left, n2 = right - mid;
  int L[n1+1], R[n2+1], i, j, k;
 
  for(i = 0 ; i < n1 ; i++){
    L[i] = a[left + i];
  }
 
  for(i = 0 ; i < n2 ; i++){
    R[i] = a[mid + i];
  }
 
  L[n1] = R[n2] = INF;
 
  for(i = 0, j = 0, k = left ; k < right ; k++){
    if(i < n1 && (j == n2 || L[i] <= R[j])){
      a[k] = L[i++];
    }else{
      ans += (n1 + n2) / 2 - i;
      a[k] = R[j++];
    }
  }
}
 
void mergesort(int a[], int left, int right){
  int mid;
 
  if(left + 1 < right){
    mid = (left + right) / 2;
    mergesort(a, left, mid);
    mergesort(a, mid, right);
    merge(a, left, mid, right);
  }
}