#include<bits/stdc++.h>
#define rep(i,n)for(int i=0;i<n;i++)
using namespace std;

int bit[200000], s, a[200000], b[200000], n;
void add(int k, int x) {
	while (k <= s) {
		bit[k] += x;
		k += k&-k;
	}
}
int sum(int k) {
	int res = 0;
	while (k > 0) {
		res += bit[k];
		k -= k&-k;
	}
	return res;
}
void compress() {
	sort(b, b + n);
	int t = unique(b, b + n) - b;
	rep(i, n)a[i] = lower_bound(b, b + t, a[i]) - b + 1;
	s = t + 1;
}
int main() {
	scanf("%d", &n);
	long long ans = 0;
	rep(i, n) {
		scanf("%d", &a[i]); b[i] = a[i];
	}
	compress();
	rep(i, n) {
		ans += i - sum(a[i]);
		add(a[i], 1);
	}
	printf("%lld\n", ans);
}
### Prompt

Generate 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<bits/stdc++.h>
#define rep(i,n)for(int i=0;i<n;i++)
using namespace std;

int bit[200000], s, a[200000], b[200000], n;
void add(int k, int x) {
	while (k <= s) {
		bit[k] += x;
		k += k&-k;
	}
}
int sum(int k) {
	int res = 0;
	while (k > 0) {
		res += bit[k];
		k -= k&-k;
	}
	return res;
}
void compress() {
	sort(b, b + n);
	int t = unique(b, b + n) - b;
	rep(i, n)a[i] = lower_bound(b, b + t, a[i]) - b + 1;
	s = t + 1;
}
int main() {
	scanf("%d", &n);
	long long ans = 0;
	rep(i, n) {
		scanf("%d", &a[i]); b[i] = a[i];
	}
	compress();
	rep(i, n) {
		ans += i - sum(a[i]);
		add(a[i], 1);
	}
	printf("%lld\n", ans);
}
```