#include <bits/stdc++.h>
using namespace std;

int n;
vector<pair<int, int>> a;
vector<int> bit;
long long ans;

int getsum(int x){ // [1, x]
	int res = 0;
	while(x > 0){
		res += bit[x];
		x -= x & -x;
	}
	return res;
}

void update(int x, int t){
	while(x <= n){
		bit[x] += t;
		x += x & -x;
	}
}

int main(){
//	cin.tie(0);
//	ios::sync_with_stdio(false);
	cin >> n;
	bit.resize(n+1);
	for(int i=0; i<n; ++i){
		int t;
		cin >> t;
		a.push_back(make_pair(t, i));
	}
	sort(a.begin(), a.end());
	for(int i=0; i<n; ++i)
		a[i].first = i+1;
	sort(a.begin(), a.end(), [](const pair<int, int> &x, const pair<int, int> &y){return x.second < y.second;});
	for(int i=n-1; i>=0; --i){
		ans += getsum(a[i].first);
		update(a[i].first, 1);
	}
//	for(int i=0; i<n; ++i)
//		cout << a[i].first << " " << a[i].second << "\n";
	cout << ans << "\n";
}

### Prompt

Please create a solution in CPP 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 <bits/stdc++.h>
using namespace std;

int n;
vector<pair<int, int>> a;
vector<int> bit;
long long ans;

int getsum(int x){ // [1, x]
	int res = 0;
	while(x > 0){
		res += bit[x];
		x -= x & -x;
	}
	return res;
}

void update(int x, int t){
	while(x <= n){
		bit[x] += t;
		x += x & -x;
	}
}

int main(){
//	cin.tie(0);
//	ios::sync_with_stdio(false);
	cin >> n;
	bit.resize(n+1);
	for(int i=0; i<n; ++i){
		int t;
		cin >> t;
		a.push_back(make_pair(t, i));
	}
	sort(a.begin(), a.end());
	for(int i=0; i<n; ++i)
		a[i].first = i+1;
	sort(a.begin(), a.end(), [](const pair<int, int> &x, const pair<int, int> &y){return x.second < y.second;});
	for(int i=n-1; i>=0; --i){
		ans += getsum(a[i].first);
		update(a[i].first, 1);
	}
//	for(int i=0; i<n; ++i)
//		cout << a[i].first << " " << a[i].second << "\n";
	cout << ans << "\n";
}

```