#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
int n;
int a[100001],b[100001];
long long suma,sumb;
long long sum[100001];
int c[200001],cnt;
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
		scanf("%d%d",&a[i],&b[i]);
		c[++cnt]=a[i],c[++cnt]=b[i];
		suma+=a[i],sumb+=b[i];
	}
	sort(c+1,c+cnt+1);
	for(int i=1;i<=n;i++)
		sum[i]=sum[i-1]+c[i];
	long long ans=min(suma,sumb);
	for(int i=1;i<=n;i++)
	{
		long long x=a[i],y=b[i];
		if(x>y) swap(x,y);
		if(lower_bound(c+1,c+cnt+1,x)-c<=n-2)
  		{
   			if(lower_bound(c+1,c+cnt+1,y)-c<=n-1) ans=min(ans,sum[n]);
   			else ans=min(ans,y+sum[n-1]);
  		}
  		else ans=min(ans,x+y+sum[n-2]);
	}
	printf("%lld",ans);
	return 0;
}
### Prompt

Create a solution in cpp for 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<cstdio>
#include<algorithm>
using namespace std;
int n;
int a[100001],b[100001];
long long suma,sumb;
long long sum[100001];
int c[200001],cnt;
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
		scanf("%d%d",&a[i],&b[i]);
		c[++cnt]=a[i],c[++cnt]=b[i];
		suma+=a[i],sumb+=b[i];
	}
	sort(c+1,c+cnt+1);
	for(int i=1;i<=n;i++)
		sum[i]=sum[i-1]+c[i];
	long long ans=min(suma,sumb);
	for(int i=1;i<=n;i++)
	{
		long long x=a[i],y=b[i];
		if(x>y) swap(x,y);
		if(lower_bound(c+1,c+cnt+1,x)-c<=n-2)
  		{
   			if(lower_bound(c+1,c+cnt+1,y)-c<=n-1) ans=min(ans,sum[n]);
   			else ans=min(ans,y+sum[n-1]);
  		}
  		else ans=min(ans,x+y+sum[n-2]);
	}
	printf("%lld",ans);
	return 0;
}
```