#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int N=200005;
int n,f1,f2,s[N];ll ans;
struct st{int x,o;}p[N];
inline bool cmp(const st&a,const st&b){return a.x<b.x;}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)scanf("%d%d",&p[i].x,&p[i+n].x),p[i].o=i,p[i+n].o=-i;
	sort(p+1,p+n+n+1,cmp);
	for(int i=1;i<=n;i++)ans+=p[i].x;
	for(int i=1;i<=n;i++)if(s[abs(p[i].o)]++)f1=1;
	for(int i=1;i<n;i++)if((p[i].o<0&&p[i+1].o>0)||(p[i].o>0&&p[i+1].o<0))f2=1;
	if(f1||!f2)printf("%lld\n",ans);
	else if(p[n].o+p[n+1].o)printf("%lld\n",ans-p[n].x+p[n+1].x);
	else printf("%lld\n",ans+min(p[n+2].x-p[n].x,p[n+1].x-p[n-1].x));
	return 0;
}

### Prompt

In CPP, your task is to solve 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;
#define ll long long
const int N=200005;
int n,f1,f2,s[N];ll ans;
struct st{int x,o;}p[N];
inline bool cmp(const st&a,const st&b){return a.x<b.x;}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)scanf("%d%d",&p[i].x,&p[i+n].x),p[i].o=i,p[i+n].o=-i;
	sort(p+1,p+n+n+1,cmp);
	for(int i=1;i<=n;i++)ans+=p[i].x;
	for(int i=1;i<=n;i++)if(s[abs(p[i].o)]++)f1=1;
	for(int i=1;i<n;i++)if((p[i].o<0&&p[i+1].o>0)||(p[i].o>0&&p[i+1].o<0))f2=1;
	if(f1||!f2)printf("%lld\n",ans);
	else if(p[n].o+p[n+1].o)printf("%lld\n",ans-p[n].x+p[n+1].x);
	else printf("%lld\n",ans+min(p[n+2].x-p[n].x,p[n+1].x-p[n-1].x));
	return 0;
}

```