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 <iostream>
#include<stdio.h>
using namespace std;
#define MAX 200000
#define mm  10000000000
typedef long long ll;
int L[MAX],R[MAX];
ll merge(int A[],int n,int left,int mid,int right)
{

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

}
ll mergeSort(int A[],int n,int left, int right)
{
    int mid;
    ll v1,v2,v3;
    if(left+1<right)
    {
        mid=(left+right)/2;
        v1=mergeSort(A,n,left,mid);
        v2=mergeSort(A,n,mid,right);
        v3=merge(A,n,left,mid,right);
        return v1+v2+v3;
    }
    return 0;
}
int main()
{
    int A[MAX],n;
    scanf("%d",&n);
    for(int i=0; i<n; i++)
    {
        scanf("%d",&A[i]);
    }
    ll ans=mergeSort(A,n,0,n);
    cout<<ans<<endl;
    return 0;
}