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

class BIT
{
public:
    vector<int> bit;
    int M;

    BIT(int M):
        bit(vector<int>(M+1, 0)), M(M) {}

    int sum(int i) {
        if (!i) return 0;
        return bit[i] + sum(i-(i&-i));
    }

    void add(int i, int x) {
        if (i > M) return;
        bit[i] += x;
        add(i+(i&-i), x);
    }
};

int main()
{
	int n;
	cin >> n;
	vector<long long> A( n );
	vector<long long> B( n );
	for( int i = 0; i < n; i++ ) {
		cin >> A[i];
		B[i] = A[i];
	}

	sort( B.begin(), B.end() );
	map<long long, int> m;
	int v = 1;
	long long last = -1;
	for ( int i = 0; i < n; i++ ) {
		if( B[i] != last ) {
			last = B[i];
			m[ B[i] ] = v;
			v++;
		}
	}
	BIT bit( m.size() + 1 );
	long long ans = 0;
	for ( int i = 0; i < n; i++ ) {
		ans += i - bit.sum( m[ A[i] ] );
		bit.add( m[ A[i] ], 1 );
	}

	cout << ans << endl;
}


### Prompt

Please provide a cpp coded solution to the problem described below:

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;

class BIT
{
public:
    vector<int> bit;
    int M;

    BIT(int M):
        bit(vector<int>(M+1, 0)), M(M) {}

    int sum(int i) {
        if (!i) return 0;
        return bit[i] + sum(i-(i&-i));
    }

    void add(int i, int x) {
        if (i > M) return;
        bit[i] += x;
        add(i+(i&-i), x);
    }
};

int main()
{
	int n;
	cin >> n;
	vector<long long> A( n );
	vector<long long> B( n );
	for( int i = 0; i < n; i++ ) {
		cin >> A[i];
		B[i] = A[i];
	}

	sort( B.begin(), B.end() );
	map<long long, int> m;
	int v = 1;
	long long last = -1;
	for ( int i = 0; i < n; i++ ) {
		if( B[i] != last ) {
			last = B[i];
			m[ B[i] ] = v;
			v++;
		}
	}
	BIT bit( m.size() + 1 );
	long long ans = 0;
	for ( int i = 0; i < n; i++ ) {
		ans += i - bit.sum( m[ A[i] ] );
		bit.add( m[ A[i] ], 1 );
	}

	cout << ans << endl;
}


```