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
#include<stdio.h>
#include<algorithm>
using namespace std;
struct node{
	int id;
	int val;
	bool operator<(node b) const
	{
		return val<b.val;
	}
};
node nn[262144];
bool vis[131072];
int main()
{
	int n;
	scanf("%d",&n);
	long long ans1=0,ans2=0;
	for(int i=0;i<(n<<1);i++)
	{
		scanf("%d",&nn[i].val);
		nn[i].id=(i>>1);
		if(i&1)
		{
			ans1+=nn[i].val;
		}
		else
		{
			ans2+=nn[i].val;
		}
	}
	sort(nn,nn+(n<<1));
	long long ans=0;
	bool ok=false;
	for(int i=0;i<n;i++)
	{
		ans+=nn[i].val;
		if(vis[nn[i].id])
		{
			ok=true;
		}
		vis[nn[i].id]=true;
	}
	if(ok)
	{
		printf("%lld\n",ans);
		return 0;
	}
	ans-=nn[n-1].val;
	if(nn[n-1].id!=nn[n].id)
	{
		ans+=nn[n].val;
	}
	else
	{
		ans-=nn[n-2].val;
		if(nn[n-2].val+nn[n+1].val<nn[n-1].val+nn[n].val)
		{
			ans+=nn[n-2].val+nn[n+1].val;
		}
		else
		{
			ans+=nn[n-1].val+nn[n].val;
		}
	}
	if(ans1<ans)
	{
		ans=ans1;
	}
	if(ans2<ans)
	{
		ans=ans2;
	}
	printf("%lld\n",ans);
	return 0;
}