#include<bits/stdc++.h>
using namespace std;
struct Info{int nu,we,qq;}a[1000010];
bool comp(const Info &a,const Info &b){return a.nu<b.nu;}
int n,b[1000010];long long ansn;bool c;
int main(){
	cin>>n;
	for (int i=1;i<=n;i++){cin>>a[i].nu>>a[i+n].nu;a[i].we=a[i+n].we=i;a[i].qq=1;a[i+n].qq=2;}
	sort(a+1,a+n+n+1,comp);
	for (int i=1;i<=n;i++){ansn+=a[i].nu;b[a[i].we]++;if (b[a[i].we]==2) c=true;}
	bool d=true,e=true;
	for (int i=2;i<=n;i++)if (a[i].qq!=a[1].qq)d=false;
	for (int i=2;i<n;i++) if (a[i].qq!=a[1].qq)e=false;
	if (c||d)cout<<ansn<<endl;
	else {
		if (a[n].we==a[n+1].we&&!e){ansn-=a[n].nu+a[n-1].nu;cout<<min(ansn+a[n-1].nu+a[n+2].nu,ansn+a[n].nu+a[n+1].nu);}
		else{cout<<ansn-a[n].nu+a[n+1].nu;}
	}
	return 0;
}
### Prompt

Please formulate a Cpp solution to 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;
struct Info{int nu,we,qq;}a[1000010];
bool comp(const Info &a,const Info &b){return a.nu<b.nu;}
int n,b[1000010];long long ansn;bool c;
int main(){
	cin>>n;
	for (int i=1;i<=n;i++){cin>>a[i].nu>>a[i+n].nu;a[i].we=a[i+n].we=i;a[i].qq=1;a[i+n].qq=2;}
	sort(a+1,a+n+n+1,comp);
	for (int i=1;i<=n;i++){ansn+=a[i].nu;b[a[i].we]++;if (b[a[i].we]==2) c=true;}
	bool d=true,e=true;
	for (int i=2;i<=n;i++)if (a[i].qq!=a[1].qq)d=false;
	for (int i=2;i<n;i++) if (a[i].qq!=a[1].qq)e=false;
	if (c||d)cout<<ansn<<endl;
	else {
		if (a[n].we==a[n+1].we&&!e){ansn-=a[n].nu+a[n-1].nu;cout<<min(ansn+a[n-1].nu+a[n+2].nu,ansn+a[n].nu+a[n+1].nu);}
		else{cout<<ansn-a[n].nu+a[n+1].nu;}
	}
	return 0;
}
```