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 <stdlib.h>
#define MAX 200000
#define INFTY 2000000000
int cnt=0;
void bubble(int [],int,int,int);
void bubbleSort(int [],int,int);

void bubble(int s[],int left,int mid,int right){
  int i,j=0,k=0;
  int x1=mid-left,x2=right-mid;
  int L[MAX]={},R[MAX]={};
  
  for(i=0;i<x1;i++){
    L[i]=s[left+i];
  }

  for(i=0;i<x2;i++){
    R[i]=s[mid+i];
  }

  L[x1]=INFTY;
  R[x2]=INFTY;

  i=0;
  j=0;
  k=0;
  
  for(k=left;k<right;k++){
    if(L[i]<=R[j]){
      s[k]=L[i++];
    }else{
      s[k]=R[j++];
      if(L[i]<INFTY) cnt=cnt+x1-i;
    }
  }
  
}

void bubbleSort(int s[],int left,int right){
  int middle;

  if(left+1<right){
    middle=(left+right+1)/2;
    bubbleSort(s,left,middle);
    bubbleSort(s,middle,right);
    bubble(s,left,middle,right);
  }
}

int main()
{
  int i,n,s[MAX];

  scanf("%d",&n);

  for(i=0;i<n;i++){
    scanf("%d",&s[i]);
    if(s[i]<1&&1000000000<s[i]) exit(8);
  }

  bubbleSort(s,0,n);

  printf("%d\n",cnt);

  return 0;
}