#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define LL long long
using namespace std;
LL n,a[100010],b[100010],c[200010],sum[200010];
LL t1=0,t2=0,ans;
int main()
{
	scanf("%lld",&n);
	for(LL i=1;i<=n;i++)
	{
		scanf("%lld %lld",&a[i],&b[i]);
		c[i]=a[i];c[n+i]=b[i];
		t1+=a[i];t2+=b[i];
	}
	sort(c+1,c+2*n+1);
	ans=min(t1,t2);
	for(LL i=1;i<=2*n;i++) sum[i]=sum[i-1]+c[i];
	for(LL i=1;i<=n;i++)
	{
		LL x=a[i],y=b[i];
		if(x>y) swap(x,y);
		if(lower_bound(c+1,c+2*n+1,x)-c<=n-2)
		{
			if(lower_bound(c+1,c+2*n+1,y)-c<=n-1) ans=min(ans,sum[n]);
			else ans=min(ans,sum[n-1]+y);
		}
		else ans=min(ans,sum[n-2]+x+y);
	}
	printf("%lld",ans);
}
### 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<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define LL long long
using namespace std;
LL n,a[100010],b[100010],c[200010],sum[200010];
LL t1=0,t2=0,ans;
int main()
{
	scanf("%lld",&n);
	for(LL i=1;i<=n;i++)
	{
		scanf("%lld %lld",&a[i],&b[i]);
		c[i]=a[i];c[n+i]=b[i];
		t1+=a[i];t2+=b[i];
	}
	sort(c+1,c+2*n+1);
	ans=min(t1,t2);
	for(LL i=1;i<=2*n;i++) sum[i]=sum[i-1]+c[i];
	for(LL i=1;i<=n;i++)
	{
		LL x=a[i],y=b[i];
		if(x>y) swap(x,y);
		if(lower_bound(c+1,c+2*n+1,x)-c<=n-2)
		{
			if(lower_bound(c+1,c+2*n+1,y)-c<=n-1) ans=min(ans,sum[n]);
			else ans=min(ans,sum[n-1]+y);
		}
		else ans=min(ans,sum[n-2]+x+y);
	}
	printf("%lld",ans);
}
```