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<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<string>
#include<algorithm>
#define N 100005
#define ll long long
using namespace std;
int n,x,y;ll ans=1e17,sum1,sum2,sum;
int fl[N],fil[N];
struct pl{int id,num;}s[2*N];
bool cmp(const pl &A,const pl &B){return A.num<B.num;}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
		scanf("%d%d",&x,&y);
		sum1+=x;sum2+=y;s[2*i-1].num=x;s[2*i-1].id=i;
		s[2*i].num=y;s[2*i].id=i;
	}
	ans=min(sum1,sum2);
	sort(s+1,s+1+2*n,cmp);
	for(int i=1;i<=n;i++) fl[s[i].id]++,sum+=s[i].num;
	for(int i=n+1;i<=2*n;i++) fil[s[i].id]+=s[i].num;
	for(int i=1;i<=n;i++){
		if(fl[s[i].id]==2) ans=min(ans,sum);
		else{
			if(i!=n) ans=min(ans,sum+fil[s[i].id]-s[n].num);
			else ans=min(ans,sum+fil[s[i].id]-s[n-1].num);
		}
	}
	printf("%lld\n",ans);
	return 0;
} 