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: Wrong Answer)

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

public class Main {

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

		int n = Integer.parseInt(br.readLine());

		String[] data = br.readLine().split(" ");

		int[] A = new int[n];
		for (int i=0; i < n; i++) {
			A[i] = Integer.parseInt(data[i]);
		}

		int h = mergeSort(A, 0, n);
		System.out.println(h);

	}

	static int merge(int[] A, int l, int m, int r) { // l:left, m:mid, r:right
		int n1 = m - l;
		int n2 = r - m;

		int max = Integer.MAX_VALUE; //bannpei
		// L = {A[l], A[l+1], A[l+2], ... , A[m-2], A[m-1], bannpei}
		int[] L = new int[n1+1];
		for (int i=0; i < n1; i++) {
			L[i] = A[l+i];
		}

		// R = {A[m], A[m+1], ... , A[r-1], bannpei}
		int[] R = new int[n2+1];
		for (int i=0; i < n2; i++) {
			R[i] = A[m+i];
		}

		// bannpei?¨????
		L[n1] = max;
		R[n2] = max;

		int i = 0;
		int j = 0;
		int count = 0;
		for (int k = l; k < r; k++) {
			if ( L[i] <= R[j] ) {
				A[k] = L[i];
				i++;
			} else {
				A[k] = R[j];
				j++;
				count += n1 - i;
			}
		}
		return count;
	}

	/**
	 * A????????????l <= index < r??¨????????¨?????????????????´????????????
	 */
	static int mergeSort(int[] A, int l, int r) {
		int count = 0;
		if ( l+1 < r ) {
			int m = (l+r) / 2;
			count += mergeSort(A, l, m);
			count += mergeSort(A, m, r);
			count += merge(A, l, m, r);
		}
		return count;
	}
}