#include<bits/stdc++.h>
using namespace std;
const int N=3e5+1;
long long n,a[N],b[N],A,B,s,v[N];
pair<int,int>p[N];
int main()
{
	cin>>n;
	for(int i=1;i<=n;i++)
	{
		cin>>a[i]>>b[i];
		p[i].first=a[i],p[i+n].first=b[i];
		p[i].second=i,p[i+n].second=i+n;
		A+=a[i],B+=b[i];
	}
	sort(p+1,p+2*n+1);
	for(int i=1;i<=n-2;i++)
		s+=p[i].first,v[p[i].second]=1;
	A=min(A,B);
	for(int i=1;i<=n;i++)
		if(v[i]+v[i+n]==2)
			A=min(A,s+p[n-1].first+p[n].first);
		else if(v[i]+v[i+n]==1)
			if(p[n-1].second==i||p[n-1].second==i+n)
				A=min(A,s+p[n].first+a[i]*(1-v[i])+b[i]*(1-v[i+n]));
			else
				A=min(A,s+p[n-1].first+a[i]*(1-v[i])+b[i]*(1-v[i+n]));
		else
			A=min(A,s+a[i]+b[i]);
	cout<<A;
	return 0;
}
### Prompt

Your task is to create a CPP solution to 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<bits/stdc++.h>
using namespace std;
const int N=3e5+1;
long long n,a[N],b[N],A,B,s,v[N];
pair<int,int>p[N];
int main()
{
	cin>>n;
	for(int i=1;i<=n;i++)
	{
		cin>>a[i]>>b[i];
		p[i].first=a[i],p[i+n].first=b[i];
		p[i].second=i,p[i+n].second=i+n;
		A+=a[i],B+=b[i];
	}
	sort(p+1,p+2*n+1);
	for(int i=1;i<=n-2;i++)
		s+=p[i].first,v[p[i].second]=1;
	A=min(A,B);
	for(int i=1;i<=n;i++)
		if(v[i]+v[i+n]==2)
			A=min(A,s+p[n-1].first+p[n].first);
		else if(v[i]+v[i+n]==1)
			if(p[n-1].second==i||p[n-1].second==i+n)
				A=min(A,s+p[n].first+a[i]*(1-v[i])+b[i]*(1-v[i+n]));
			else
				A=min(A,s+p[n-1].first+a[i]*(1-v[i])+b[i]*(1-v[i+n]));
		else
			A=min(A,s+a[i]+b[i]);
	cout<<A;
	return 0;
}
```