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.FileInputStream;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;

public class Main {
	public static int cnt = 0;
	public static void main(String[] args) throws IOException {
		//System.setIn(new FileInputStream("./input.txt"));
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(br.readLine());
		/*ArrayList<Integer> a = new ArrayList<Integer>();
		String[] s = br.readLine().split(" ");
		for(int i = 0; i < n; i++) {
			a.add(Integer.parseInt(s[i]));
		}*/
		int[] s = Arrays.stream(br.readLine().split(" ")).mapToInt(Integer::parseInt).toArray();
		
		mergeSort(s, 0, n);

		System.out.println(cnt);
	}
	public static void printArray(int[] s){
		for(int i = 0; i < s.length; ++i) {
			System.out.print(" " + s[i]);
		}
		System.out.println(": " + cnt);
	}
	public static void mergeSort(int[] s, int left, int right) {
		if(left + 1 < right) {
			int mid = (left + right) / 2;
			mergeSort(s, left, mid);
			mergeSort(s, mid, right);
			merge(s, left, mid, right);
			//printArray(s);
		}
	}
	public static void merge(int[] s, int left, int mid, int right) {
		int[] s1 = Arrays.copyOfRange(s, left, mid + 1);
		int[] s2 = Arrays.copyOfRange(s, mid, right + 1);
		
		s1[mid - left] = Integer.MAX_VALUE;
		s2[right - mid] = Integer.MAX_VALUE;
		int i = 0;
		int j = 0;
		for(int k = left; k < right; ++k) {
			if(s1[i] <= s2[j]){
				s[k] = s1[i];
				if(k < left + i){
					cnt += left + i - k;
				}
				i += 1;
				
			}else{
				s[k] = s2[j];
				if(k < mid + j){
					cnt += mid + j - k;
				}
				j += 1;
				
			}
		}
		
	}
	
}