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<bits/stdc++.h>
#define INF 1001001001

long long int cnt;

void merge(std::vector<int>&A, int left, int mid, int right){
    int n1 = mid - left;
    int n2 = right - mid;
    std::vector<int> L(n1+1);
    std::vector<int> R(n2+1);
    for(int i = 0; i < n1; i++){
        L[i] = A[left + i];
    }
    for(int i = 0; i < n2; i++){
        R[i] = A[mid + i];
    }
    L[n1] = INF;
    R[n2] = INF;
    int i = 0;
    int j = 0;

    for(int k = left; k < right; k++){
        if(L[i] <= R[j]){
            A[k] = L[i];
            i++;
        }else{
            A[k] = R[j];
            j++;
            cnt += n1 - i;
        }
        
    }
    
}

void mergeSort(std::vector<int>&A, int left, int right){
    if(left + 1 < right){
        int mid = (left + right) / 2;
        mergeSort(A, left, mid);
        mergeSort(A, mid, right);
        merge(A, left, mid, right);
    }
}

int main(void){
    int n;
    cnt = 0;
    std::cin >> n;
    std::vector<int> A(n);

    for(int i = 0; i < n; i++){
        std::cin >> A[i];
    }
    

    mergeSort(A, 0, n);

    
    std::cout << cnt << std::endl;
    
    return 0;
}