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<cstring>
#include<algorithm>
#include<cstdlib>
#define N 200010
#define ll long long
using namespace std;
struct node{
	ll a,b;
	int c;
	bool operator <(const node d)const{return a<d.a;}
}v[N];
bool vis[N];
int s[N];
int main()
{
	int n,t=0;
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
		scanf("%lld%lld",&v[t+1].a,&v[t+2].a);
		v[t+1].b=v[t+2].b=i;
		v[t+1].c=0,v[t+2].c=1;
		t+=2;
	}
	sort(v+1,v+t+1);
    long long res=0;
    bool have=false;
    for(int i=1;i<=n;i++)
	{
		s[v[i].c]++;
		res+=v[i].a;
		if(vis[v[i].b]) have=true;
		else vis[v[i].b]=true;
	}
	if(have || s[0]==n || s[1]==n){printf("%lld\n",res);return 0;}
	res+=v[n+1].a-v[n].a;
	s[v[n].c]--,s[v[n+1].c]++;
	if(s[0]==n || s[1]==n){printf("%lld\n",res);return 0;}
	if(v[n].b==v[n+1].b) res+=min(v[n+2].a-v[n+1].a,v[n].a-v[n-1].a);
	printf("%lld\n",res);
	return 0;
}