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>
using namespace std;

#define MAX 200000
#define SENTINEL 2000000000
typedef long long llong;

int L[MAX / 2 + 2], R[MAX / 2 + 2];

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

llong mergeSort(int A[], int n, int left, int right)
{
    int mid;
    llong 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;
    }
    else
    {
        return 0;
    }
}

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

    cin >> n;
    for (i = 0; i < n; i++)
        cin >> A[i];

    llong ans = mergeSort(A, n, 0, n);
    cout << ans << endl;

    return 0;
}