#include<bits/stdc++.h>
using namespace std;
struct val
{
	int v,id,type;
	bool operator<(const val &b)const{return v<b.v;}
	void read(int ID,int op){scanf("%d",&v);id=ID;type=op;}
}a[300001];
int n,vis[300001];
int main()
{
	scanf("%d",&n);
	for(int i=0;i<n;i++)a[i].read(i,1),a[i+n].read(i,-1);
	sort(a,a+2*n);
	long long ans=0;
	int x=0,y=0;
	for(int i=0;i<n;i++)
	{
		if(i&&a[i].type*a[i-1].type<0)y=1;
		if(vis[a[i].id])x=1;
		ans+=a[i].v;
		vis[a[i].id]=1;
	}
	if(x||!y)printf("%lld\n",ans);
	else printf("%lld\n",a[n-1].id==a[n].id?ans+min(a[n].v-a[n-2].v,a[n+1].v-a[n-1].v):ans-a[n-1].v+a[n].v);
}

### Prompt

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

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;
struct val
{
	int v,id,type;
	bool operator<(const val &b)const{return v<b.v;}
	void read(int ID,int op){scanf("%d",&v);id=ID;type=op;}
}a[300001];
int n,vis[300001];
int main()
{
	scanf("%d",&n);
	for(int i=0;i<n;i++)a[i].read(i,1),a[i+n].read(i,-1);
	sort(a,a+2*n);
	long long ans=0;
	int x=0,y=0;
	for(int i=0;i<n;i++)
	{
		if(i&&a[i].type*a[i-1].type<0)y=1;
		if(vis[a[i].id])x=1;
		ans+=a[i].v;
		vis[a[i].id]=1;
	}
	if(x||!y)printf("%lld\n",ans);
	else printf("%lld\n",a[n-1].id==a[n].id?ans+min(a[n].v-a[n-2].v,a[n+1].v-a[n-1].v):ans-a[n-1].v+a[n].v);
}

```