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: Wrong Answer)

//逆序数 
#include <iostream>

using namespace std;


int merge(int *arr,int left,int m,int right) {
	int n1 = m -left;
	int n2 = right -m;
	int *L = new int[n1];
	int *R = new int[n2];
	int index1 = 0,index2  =0;
	int cnt = 0;
	for (int i = left;i<right;i++) {
		if (i < m) {
			L[index1++] = arr[i];
		} else {
			R[index2++] = arr[i];
		}
	}
	index1 = 0;
	index2 = 0;
	for (int i = left;i<right;i++) {
		if (index1 >= n1) {
			arr[i] = R[index2++];
		} else if (index2 >= n2) {
			arr[i] = L[index1++];
		} else {
			if (L[index1] <= R[index2]) {
				arr[i] = L[index1++];
			} else {
				cnt += n1 - index1; 
				arr[i] = R[index2++];
			}
		}
	}
	return cnt;
}

//包括left但不包括right 
int mergeSort(int *arr,int left,int right) {
	if (right - left > 1) {
		int m = (left + right) / 2;
		int a = mergeSort(arr,left,m);
		int b = mergeSort(arr,m,right);
		int c = merge(arr,left,m,right);
		return a +  b + c;
	} else {
		return 0;
	}
	
}
int main() {
	int n;
	cin >> n;
	int *arr = new int[n];
	for (int i=0;i<n;i++) {
		cin >> arr[i];
	}
	int x = mergeSort(arr,0,n);
	cout << x << endl;
}