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 <unistd.h>
#include <stdlib.h>

#define MAX 200000 //数列の長さn
#define AI 1000000001

int cnt;

void merge (int A[], int left, int mid, int right){
  int n1, n2, i, j, k;
  int L[MAX] = {0}, R[MAX] = {0};
  n1 = mid - left;
  n2 = right - mid;
  // L=n1+1;
  // R=n2+1;
  for(i = 0; i <= n1 - 1; i++){
    L[i] = A[left + i];
  }
  for(j = 0; j <= n2 - 1; j++){
    R[j] = A[mid + j];
  }
  L[n1] = MAX;
  R[n2] = MAX;
  i = 0;
  j = 0;
  for(k = left; k <= right - 1; k++){
   if(L[i] <= R[j]){
      A[k] = L[i];
      i++;
    }
    else{
      A[k] = R[j];
      j++;
      cnt += (n1 - i);
    }
  }
  return;
}

void mergeSort (int A[], int left, int right){
  int i, mid;
  if (left + 1 < right){
    mid = (left + right) / 2;
    mergeSort(A, left, mid);
    mergeSort(A, mid, right);
    merge(A, left, mid, right);
  }
  return;
}


int main (void){
  int A[MAX];
  int n, i;
  scanf("%d", &n);
  for (i = 0; i < n; i++){
    scanf("%d", &A[i]);
  }
  cnt = 0;
  mergeSort(A, 0, n);
  // for(i=0;i<n;i++){
  //     printf("%lld",A[i]);
  //     if(i<n-1)printf(" ");
  //   }
    // printf("\n");
    printf("%d\n", cnt);
      return 0;
}