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>
  
int bubble_sort(int* A, int N){
  int i,j, cnt = 0,flag = 0;
  for(i = 0; i < N; i++)
  {
      flag = 0;
      for(j = N-1; j >= i+1; j--)
      {
          if(A[j] < A[j-1]){
              int tmp = A[j];
              A[j] = A[j-1];
              A[j-1] = tmp;
              cnt++;
              flag = 1;
          }
          if(j == i && flag == 0)
          {
              return cnt;
          }
      }
  }
  return cnt;
}
  
int main(){
  int i,N,data[200000];
  scanf("%d",&N);
  for(i = 0; i < N; i++) scanf("%d",&data[i]);
  printf("%d\n",bubble_sort(data,N));
  return 0;
}