#include<iostream>
#include<vector>

long long int mergecount(std::vector<int> &a) {
	long long int count = 0;
	int n = a.size();
	if (n > 1) {
		std::vector<int> b(a.begin(), a.begin() + n / 2);
		std::vector<int> c(a.begin() + n / 2, a.end());
		count += mergecount(b);
		count += mergecount(c);
		for (int i = 0, j = 0, k = 0; i < n; ++i)
			if (k == c.size())       a[i] = b[j++];
			else if (j == b.size())  a[i] = c[k++];
			else if (b[j] <= c[k])   a[i] = b[j++];
			else { a[i] = c[k++]; count += n / 2 - j; }
	}
	return count;
}

int main() {
	int n;
	std::cin >> n;
	std::vector<int> a(n,0);

	
	for (int i = 0; i<n; i++) {
		std::cin >> a[i];
	}
	std::cout << mergecount(a) << std::endl;
		return 0;
}
### Prompt

Your task is to create a CPP solution to the following problem:

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.

Constraints

* $ 1 \leq n \leq 200,000$
* $ 0 \leq a_i \leq 10^9$
* $a_i$ are all different

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.

Examples

Input

5
3 5 2 1 4


Output

6


Input

3
3 1 2


Output

2

### Response

```cpp
#include<iostream>
#include<vector>

long long int mergecount(std::vector<int> &a) {
	long long int count = 0;
	int n = a.size();
	if (n > 1) {
		std::vector<int> b(a.begin(), a.begin() + n / 2);
		std::vector<int> c(a.begin() + n / 2, a.end());
		count += mergecount(b);
		count += mergecount(c);
		for (int i = 0, j = 0, k = 0; i < n; ++i)
			if (k == c.size())       a[i] = b[j++];
			else if (j == b.size())  a[i] = c[k++];
			else if (b[j] <= c[k])   a[i] = b[j++];
			else { a[i] = c[k++]; count += n / 2 - j; }
	}
	return count;
}

int main() {
	int n;
	std::cin >> n;
	std::vector<int> a(n,0);

	
	for (int i = 0; i<n; i++) {
		std::cin >> a[i];
	}
	std::cout << mergecount(a) << std::endl;
		return 0;
}
```