#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<vector>
using namespace std;
typedef long long LL;
typedef pair<LL,LL> PI;
const LL N=200005;
LL n;
LL a[N],b[N];
LL ans=0,sum=0;
PI c[N];
LL tot;
bool cmp (PI x,PI y)	{return x.first>y.first;}
bool in[N];
int main()
{
	scanf("%lld",&n);
	for (LL u=1;u<=n;u++) 
	{
		scanf("%lld%lld",&a[u],&b[u]);
		ans=ans+a[u]+b[u];sum=sum+a[u];
		c[++tot]={a[u],u};c[++tot]={b[u],u};
	}
	sort(c+1,c+1+tot,cmp);
	LL mx=0;
	mx=max(sum,ans-sum);sum=0;
	bool tf=false;
	memset(in,false,sizeof(in));
	for (LL u=1;u<=n;u++)
	{
		if (in[c[u].second]) tf=true;
		sum=sum+c[u].first;
		in[c[u].second]=true;
	}
	if (tf) mx=max(mx,sum);
	else
	{
		for (LL i=n+1;i<=tot;i++)
		{
			if (c[i].second!=c[n].second) mx=max(mx,sum+c[i].first-c[n].first);
			else mx=max(mx,sum+c[i].first-c[n-1].first);
		}
	}
	printf("%lld\n",ans-mx);
	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<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<vector>
using namespace std;
typedef long long LL;
typedef pair<LL,LL> PI;
const LL N=200005;
LL n;
LL a[N],b[N];
LL ans=0,sum=0;
PI c[N];
LL tot;
bool cmp (PI x,PI y)	{return x.first>y.first;}
bool in[N];
int main()
{
	scanf("%lld",&n);
	for (LL u=1;u<=n;u++) 
	{
		scanf("%lld%lld",&a[u],&b[u]);
		ans=ans+a[u]+b[u];sum=sum+a[u];
		c[++tot]={a[u],u};c[++tot]={b[u],u};
	}
	sort(c+1,c+1+tot,cmp);
	LL mx=0;
	mx=max(sum,ans-sum);sum=0;
	bool tf=false;
	memset(in,false,sizeof(in));
	for (LL u=1;u<=n;u++)
	{
		if (in[c[u].second]) tf=true;
		sum=sum+c[u].first;
		in[c[u].second]=true;
	}
	if (tf) mx=max(mx,sum);
	else
	{
		for (LL i=n+1;i<=tot;i++)
		{
			if (c[i].second!=c[n].second) mx=max(mx,sum+c[i].first-c[n].first);
			else mx=max(mx,sum+c[i].first-c[n-1].first);
		}
	}
	printf("%lld\n",ans-mx);
	return 0;
}
```