#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
struct data
{
	int w,id;
	friend bool operator < (data a,data b){return a.w<b.w;}
}d[200020];
int n,a[100010],b[100010],cnt1,cnt2;
long long ans,sum;
bool usea[100010],useb[100010];
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;++i)
	{
		scanf("%d %d",&a[i],&b[i]);
		d[i*2-1]=data{a[i],i};
		d[i*2]=data{b[i],-i};
	}
	sort(d+1,d+1+2*n);
	int cnt1=0,cnt2=0;
	for(int i=1;i<=n;++i)
	{
		sum+=d[i].w;
		if(d[i].id>0)usea[d[i].id]=1,++cnt1;
		else useb[-d[i].id]=1,++cnt2;
	}
	bool f=0;
	for(int i=1;i<=n;++i)if(usea[i]&&useb[i])f=1;
	if(cnt1==n||cnt2==n||f)return printf("%lld",sum),0;
	int mx=0;
	for(int i=1;i<=n;++i)if(d[i].id!=-d[n+1].id)mx=max(mx,d[i].w);
	ans=sum+d[n+1].w-mx;
	mx=0;
	for(int i=1;i<=n;++i)if(d[i].id!=-d[n+2].id)mx=max(mx,d[i].w);
	ans=min(ans,sum+d[n+2].w-mx);
	printf("%lld",ans);
	return 0;
}

### Prompt

Please create a solution in Cpp 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<cstring>
#include<algorithm>
using namespace std;
struct data
{
	int w,id;
	friend bool operator < (data a,data b){return a.w<b.w;}
}d[200020];
int n,a[100010],b[100010],cnt1,cnt2;
long long ans,sum;
bool usea[100010],useb[100010];
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;++i)
	{
		scanf("%d %d",&a[i],&b[i]);
		d[i*2-1]=data{a[i],i};
		d[i*2]=data{b[i],-i};
	}
	sort(d+1,d+1+2*n);
	int cnt1=0,cnt2=0;
	for(int i=1;i<=n;++i)
	{
		sum+=d[i].w;
		if(d[i].id>0)usea[d[i].id]=1,++cnt1;
		else useb[-d[i].id]=1,++cnt2;
	}
	bool f=0;
	for(int i=1;i<=n;++i)if(usea[i]&&useb[i])f=1;
	if(cnt1==n||cnt2==n||f)return printf("%lld",sum),0;
	int mx=0;
	for(int i=1;i<=n;++i)if(d[i].id!=-d[n+1].id)mx=max(mx,d[i].w);
	ans=sum+d[n+1].w-mx;
	mx=0;
	for(int i=1;i<=n;++i)if(d[i].id!=-d[n+2].id)mx=max(mx,d[i].w);
	ans=min(ans,sum+d[n+2].w-mx);
	printf("%lld",ans);
	return 0;
}

```