#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
int n;
vector<pair<long,int> >a;
int cnt[1<<17];
int main()
{
	cin>>n;
	for(int i=0;i<2*n;i++)
	{
		int x;cin>>x;
		a.push_back(make_pair(x,i));
	}
	int f=3,u=0;
	sort(a.begin(),a.end());
	long ans=0;
	for(int i=0;i<n;i++)
	{
		ans+=a[i].first;
		u+=cnt[a[i].second/2]++;
		f&=2-a[i].second%2;
	}
	if(f)cout<<ans<<endl;
	else if(u)cout<<ans<<endl;
	else
	{
		if(a[n-1].second/2!=a[n].second/2)
		{
			cout<<ans-a[n-1].first+a[n].first<<endl;
		}
		else
		{
			cout<<ans+min(a[n+1].first-a[n-1].first,a[n].first-a[n-2].first)<<endl;
		}
	}
}
### Prompt

In cpp, your task is to solve the following problem:

We have a directed weighted graph with N vertices. Each vertex has two integers written on it, and the integers written on Vertex i are A_i and B_i.

In this graph, there is an edge from Vertex x to Vertex y for all pairs 1 \leq x,y \leq N, and its weight is {\rm min}(A_x,B_y).

We will consider a directed cycle in this graph that visits every vertex exactly once. Find the minimum total weight of the edges in such a cycle.

Constraints

* 2 \leq N \leq 10^5
* 1 \leq A_i \leq 10^9
* 1 \leq B_i \leq 10^9
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N
A_1 B_1
A_2 B_2
:
A_N B_N


Output

Print the minimum total weight of the edges in such a cycle.

Examples

Input

3
1 5
4 2
6 3


Output

7


Input

4
1 5
2 6
3 7
4 8


Output

10


Input

6
19 92
64 64
78 48
57 33
73 6
95 73


Output

227

### Response

```cpp
#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
int n;
vector<pair<long,int> >a;
int cnt[1<<17];
int main()
{
	cin>>n;
	for(int i=0;i<2*n;i++)
	{
		int x;cin>>x;
		a.push_back(make_pair(x,i));
	}
	int f=3,u=0;
	sort(a.begin(),a.end());
	long ans=0;
	for(int i=0;i<n;i++)
	{
		ans+=a[i].first;
		u+=cnt[a[i].second/2]++;
		f&=2-a[i].second%2;
	}
	if(f)cout<<ans<<endl;
	else if(u)cout<<ans<<endl;
	else
	{
		if(a[n-1].second/2!=a[n].second/2)
		{
			cout<<ans-a[n-1].first+a[n].first<<endl;
		}
		else
		{
			cout<<ans+min(a[n+1].first-a[n-1].first,a[n].first-a[n-2].first)<<endl;
		}
	}
}
```