#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
int a[100010],b[100010];
int tmp[200010];
long long f[200010];
int main()
{
	int n;scanf("%d",&n);
	long long ans1=0,ans2=0;
	for(int i=1;i<=n;i++)
	{
		scanf("%d %d",&a[i],&b[i]);
		tmp[i*2-1]=a[i];tmp[i*2]=b[i];
		ans1+=a[i];ans2+=b[i];
	}
	sort(tmp+1,tmp+n*2+1);
	for(int i=1;i<=n*2;i++)
		f[i]=f[i-1]+tmp[i];
	long long ans=min(ans1,ans2);
	for(int i=1;i<=n;i++)
	{
		int pos1=lower_bound(tmp+1,tmp+n*2+1,a[i])-tmp;
		int pos2=lower_bound(tmp+1,tmp+n*2+1,b[i])-tmp;
		if(pos1>pos2)swap(pos1,pos2);
		if(pos2<=n)ans=min(ans,f[n]);
		else if(pos1<n)ans=min(ans,f[n-1]+tmp[pos2]);
		else ans=min(ans,f[n-2]+tmp[pos1]+tmp[pos2]);
	}
	printf("%lld\n",ans);
	return 0;
}
### Prompt

Please formulate 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<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
int a[100010],b[100010];
int tmp[200010];
long long f[200010];
int main()
{
	int n;scanf("%d",&n);
	long long ans1=0,ans2=0;
	for(int i=1;i<=n;i++)
	{
		scanf("%d %d",&a[i],&b[i]);
		tmp[i*2-1]=a[i];tmp[i*2]=b[i];
		ans1+=a[i];ans2+=b[i];
	}
	sort(tmp+1,tmp+n*2+1);
	for(int i=1;i<=n*2;i++)
		f[i]=f[i-1]+tmp[i];
	long long ans=min(ans1,ans2);
	for(int i=1;i<=n;i++)
	{
		int pos1=lower_bound(tmp+1,tmp+n*2+1,a[i])-tmp;
		int pos2=lower_bound(tmp+1,tmp+n*2+1,b[i])-tmp;
		if(pos1>pos2)swap(pos1,pos2);
		if(pos2<=n)ans=min(ans,f[n]);
		else if(pos1<n)ans=min(ans,f[n-1]+tmp[pos2]);
		else ans=min(ans,f[n-2]+tmp[pos1]+tmp[pos2]);
	}
	printf("%lld\n",ans);
	return 0;
}
```