#include<bits/stdc++.h>
using namespace std;

int n,cnt,x,y,nw,hd[100010];
long long sA,sB,sum,ans;
pair<int,int> A[200010];

int main()
{
	scanf("%d",&n),sA=sB=0,cnt=0;
	for (int i=1; i<=n; i++) 
	{
		scanf("%d%d",&x,&y),sA+=x,sB+=y;
		A[++cnt]=make_pair(x,i);
		A[++cnt]=make_pair(y,i);
	}
	sum=0,ans=min(sA,sB);
	sort(A+1,A+1+cnt),nw=0;
	memset(hd,0,sizeof(hd));
	for (int i=1; i<=n; i++)
	{
		if (!hd[A[i].second]) hd[A[i].second]=i,nw++;
		sum+=A[i].first;
	}
	if (nw==n)
	{
		for (int i=n+1; i<=2*n; i++)
			if (A[i].second==A[n].second) ans=min(ans,sum-A[n-1].first+A[i].first);
			else ans=min(ans,sum-A[n].first+A[i].first);
	} else ans=min(ans,sum);
	printf("%lld\n",ans);
	return 0;
}
### Prompt

Generate 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<bits/stdc++.h>
using namespace std;

int n,cnt,x,y,nw,hd[100010];
long long sA,sB,sum,ans;
pair<int,int> A[200010];

int main()
{
	scanf("%d",&n),sA=sB=0,cnt=0;
	for (int i=1; i<=n; i++) 
	{
		scanf("%d%d",&x,&y),sA+=x,sB+=y;
		A[++cnt]=make_pair(x,i);
		A[++cnt]=make_pair(y,i);
	}
	sum=0,ans=min(sA,sB);
	sort(A+1,A+1+cnt),nw=0;
	memset(hd,0,sizeof(hd));
	for (int i=1; i<=n; i++)
	{
		if (!hd[A[i].second]) hd[A[i].second]=i,nw++;
		sum+=A[i].first;
	}
	if (nw==n)
	{
		for (int i=n+1; i<=2*n; i++)
			if (A[i].second==A[n].second) ans=min(ans,sum-A[n-1].first+A[i].first);
			else ans=min(ans,sum-A[n].first+A[i].first);
	} else ans=min(ans,sum);
	printf("%lld\n",ans);
	return 0;
}
```