#include <cstdio>
#include <vector>
using namespace std;

long long c;

void merge(int left, int mid, int right, vector<int>& A) {
	vector<int> L, R;
	int n1=mid-left;
	int n2=right-mid;
	int i,j,k;

	L.resize(n1+1);
	R.resize(n2+1);

	for(i=0; i<n1; ++i) {
		L[i]=A[left+i];
	}
	for(i=0; i<n2; ++i) {
		R[i]=A[mid+i];
	}
	i=j=0;
	L[n1]=1000000001;
	R[n2]=1000000001;
	for(k=left; k<right; ++k) {
		if(L[i]>R[j]) {
			c+=(long long)n1-(long long)i;
			A[k]=R[j];
			++j;
		} else {
			A[k]=L[i];
			++i;
		}
	}
	return;
}

void mergeSort(int left, int right, vector<int>& A) {
	if(left+1<right) {
		int mid=left+(right-left)/2;
		mergeSort(left,mid,A);
		mergeSort(mid,right,A);
		merge(left,mid,right,A);
	}
	return;
}

int main() {
	int n;
	vector<int> A;
	scanf("%d",&n);
	c=0;
	A.resize(n);
	for(int i=0; i<n; ++i) {
		scanf("%d",&A[i]);
	}
	mergeSort(0,n,A);
	printf("%lld\n",c);
	return 0;
}
### Prompt

Construct a cpp code solution to the problem outlined:

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 <cstdio>
#include <vector>
using namespace std;

long long c;

void merge(int left, int mid, int right, vector<int>& A) {
	vector<int> L, R;
	int n1=mid-left;
	int n2=right-mid;
	int i,j,k;

	L.resize(n1+1);
	R.resize(n2+1);

	for(i=0; i<n1; ++i) {
		L[i]=A[left+i];
	}
	for(i=0; i<n2; ++i) {
		R[i]=A[mid+i];
	}
	i=j=0;
	L[n1]=1000000001;
	R[n2]=1000000001;
	for(k=left; k<right; ++k) {
		if(L[i]>R[j]) {
			c+=(long long)n1-(long long)i;
			A[k]=R[j];
			++j;
		} else {
			A[k]=L[i];
			++i;
		}
	}
	return;
}

void mergeSort(int left, int right, vector<int>& A) {
	if(left+1<right) {
		int mid=left+(right-left)/2;
		mergeSort(left,mid,A);
		mergeSort(mid,right,A);
		merge(left,mid,right,A);
	}
	return;
}

int main() {
	int n;
	vector<int> A;
	scanf("%d",&n);
	c=0;
	A.resize(n);
	for(int i=0; i<n; ++i) {
		scanf("%d",&A[i]);
	}
	mergeSort(0,n,A);
	printf("%lld\n",c);
	return 0;
}
```