#include <iostream>
#include <vector>
#include <cassert>
using namespace std;
typedef long long LL;

int n;
vector<int> a;
vector<int> w;

LL merge_count(int l, int r) {
	if (r - l <= 1)
		return 0;
	else if (r - l == 2) {
		if (a[l] <= a[r-1])
			return 0;
		else {
			swap(a[l], a[r-1]);
			return 1;
		}
	}
	else {
		int m = (l + r) / 2;
		LL cl = merge_count(l, m);
		LL cr = merge_count(m, r);
		LL c = 0;
		int i=l, j=m;
		int k = l;
		while (i < m && j < r) {
			if (a[i] <= a[j])
				w[k++] = a[i++];
			else {
				w[k++] = a[j++];
				c += m-i;	//整列されているので、a[i] > a[j]のときa[i]からa[m-1]まで全てa[j]より大。
			}
		}
		while (i < m)
			w[k++] = a[i++];
		while (j < r)
			w[k++] = a[j++];
		assert(k == r);
		copy(w.begin() + l, w.begin() + r, a.begin() + l);
		return cl + cr + c;
	}
}

int main() {
	cin >> n;
	a.resize(n);
	for (int i = 0; i < n; i++)
		cin >> a[i];
	w.resize(n);
	cout << merge_count(0, n) << endl;
}

### Prompt

Create a solution in Cpp for 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>
#include <cassert>
using namespace std;
typedef long long LL;

int n;
vector<int> a;
vector<int> w;

LL merge_count(int l, int r) {
	if (r - l <= 1)
		return 0;
	else if (r - l == 2) {
		if (a[l] <= a[r-1])
			return 0;
		else {
			swap(a[l], a[r-1]);
			return 1;
		}
	}
	else {
		int m = (l + r) / 2;
		LL cl = merge_count(l, m);
		LL cr = merge_count(m, r);
		LL c = 0;
		int i=l, j=m;
		int k = l;
		while (i < m && j < r) {
			if (a[i] <= a[j])
				w[k++] = a[i++];
			else {
				w[k++] = a[j++];
				c += m-i;	//整列されているので、a[i] > a[j]のときa[i]からa[m-1]まで全てa[j]より大。
			}
		}
		while (i < m)
			w[k++] = a[i++];
		while (j < r)
			w[k++] = a[j++];
		assert(k == r);
		copy(w.begin() + l, w.begin() + r, a.begin() + l);
		return cl + cr + c;
	}
}

int main() {
	cin >> n;
	a.resize(n);
	for (int i = 0; i < n; i++)
		cin >> a[i];
	w.resize(n);
	cout << merge_count(0, n) << endl;
}

```