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<cstdio>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
int vis[101010];
priority_queue< pair<int,int> >q1,q2;
int main()
{
	int i,n,num,x,y;
	pair<int,int>tmp,tmp2;
	long long sum=0,summ,sum2,ans=0;
	scanf("%d",&n);
	for(i=1;i<=n;i++)
	{
		scanf("%d%d",&x,&y);
		q1.push(make_pair(-x,i));
		q2.push(make_pair(y,i));
		sum+=y;
		ans+=x;
	}
	summ=0,num=0,sum2=0;
	ans=min(ans,sum);
//	printf("%lld\n",sum);
	for(i=1;i<n;i++)
	{
		tmp=q1.top();
		summ+=-tmp.first;
		if(vis[tmp.second]==0)
			vis[tmp.second]=1,num++;
		tmp2=q2.top();
		sum2+=tmp2.first;
		if(vis[tmp2.second]==0)
			vis[tmp2.second]=1,num++;
		q1.pop(),q2.pop();
//		printf("%d %d %lld %lld\n",i,num,summ,sum2);
		if(num>i)
			ans=min(ans,sum-sum2+summ);
		else
		{
			if(!q1.empty())
				ans=min(ans,sum-sum2+summ-q1.top().first+tmp.first);
			if(!q2.empty())
				ans=min(ans,sum-sum2+summ-q2.top().first+tmp2.first);
		}
	}
	printf("%lld\n",ans);
	//freopen(".in","r",stdin);
	//freopen(".out","w",stdout);
 	return 0;
}
