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<string.h>
#include<stdlib.h>
#include<math.h>
#include<limits.h>
int n;
long count=0;

void merge(int *A,int left,int mid,int right){
  int *L,*R;
  L=malloc(sizeof(int)*(mid-left+1));
  R=malloc(sizeof(int)*(right-mid+1));
  
  int i;
  for(i=0;i<mid-left;i++) L[i]=A[left+i];
  L[mid-left]=INT_MAX;
  for(i=0;i<right-mid;i++) R[i]=A[mid+i];
  R[right-mid]=INT_MAX;//どの不等式にもかつ数字を、配列の最後尾に入れる
  
  int l,r;
  l=0;
  r=0;
  for(i=0;i<right-left;i++){
    if(L[l] <= R[r]){
      A[left+i]=L[l];
      l+=1;
     }
    else{
      count+=mid-left-l;
      A[left+i]=R[r];
      r+=1;
    }
  }
  free(L);
  free(R);
}




void mergesort(int *A,int left,int right){
  if(left +1 < right){//整列させる要素が２つ以上ある場合は、分割する
    int mid;
    mid=(left+right)/2;
    mergesort(A,left,mid);
    mergesort(A,mid,right);
    merge(A,left,mid,right);
  }
}
//A[left]からA[right-1]まで整列させる



int main(){
  int *A;
  scanf("%d",&n);
  A=malloc(n*sizeof(int));
  int i;
  for(i=0;i<n;i++) scanf("%d",&A[i]);

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


//どの不等式にも勝つ数字を入れたいときは、INT_MAX。limits.hをインクルードする必要あり