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 JAVA (judge verdict: Accepted)

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

class Main {
    static long inversion = 0;

    static void merge(int[] A, int left, int mid, int right) {
        int l1 = mid - left;
        int l2 = right - mid;

        int[] L = new int[l1 + 1];
        int[] R = new int[l2 + 1];

        for (int i = 0; i < l1; i++)
            L[i] = A[left + i];

        for (int i = 0; i < l2; i++)
            R[i] = A[mid + i];

        L[l1] = R[l2] = 1000000001; // Sの要素より確実に大きいInfのようなものを番兵として扱う

        int i = 0;
        int j = 0;

        for (int k = left; k < right; k++) {

            if (L[i] <= R[j]) {
                A[k] = L[i];
                i++;
            } else {
                inversion += (l1 - i);
                A[k] = R[j];
                j++;
            }

        }
    }

    static void mergeSort(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);
        }
    }

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

        int n = Integer.parseInt(br.readLine());
        String[] input = br.readLine().split(" ");
        int[] arr = new int[n];

        for (int i = 0; i < n; i++) {
            arr[i] = Integer.parseInt(input[i]);
        }

        mergeSort(arr, 0, n);

        System.out.println(inversion);
    }
}