#include<cstdio>
#include<algorithm>
#define int long long
using namespace std;
struct node{
	int val,id;
	bool operator<(node x)const{
		return val<x.val;
	}
}a[200100],b[200100],c[200100];
int sum[200100];
int n,suma,sumb,ans;
signed main(){
	scanf("%lld",&n);
	for(int i=1;i<=n;i++){
		scanf("%lld%lld",&a[i].val,&b[i].val);
		c[2*i-1].val=a[i].val,c[2*i].val=b[i].val;
		c[2*i-1].id=2*i-1,c[2*i].id=2*i;
		suma+=a[i].val,sumb+=b[i].val;
	}
	ans=min(suma,sumb);
	sort(c+1,c+2*n+1);
	for(int i=1;i<=2*n;i++){
		sum[i]=sum[i-1]+c[i].val;
		if(c[i].id%2==1)a[(c[i].id+1)/2].id=i;
		else b[c[i].id/2].id=i;
	}
	for(int i=1;i<=n;i++){
		int x;
		if(a[i].id<=n&&b[i].id<=n)x=sum[n];
		else if(a[i].id<=n-1&&b[i].id>n)x=sum[n-1]+c[b[i].id].val;
		else if(b[i].id<=n-1&&a[i].id>n)x=sum[n-1]+c[a[i].id].val;
		else x=sum[n-2]+c[a[i].id].val+c[b[i].id].val;
		ans=min(ans,x);
	}
	printf("%lld\n",ans);
}
### Prompt

Your task is to create 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<cstdio>
#include<algorithm>
#define int long long
using namespace std;
struct node{
	int val,id;
	bool operator<(node x)const{
		return val<x.val;
	}
}a[200100],b[200100],c[200100];
int sum[200100];
int n,suma,sumb,ans;
signed main(){
	scanf("%lld",&n);
	for(int i=1;i<=n;i++){
		scanf("%lld%lld",&a[i].val,&b[i].val);
		c[2*i-1].val=a[i].val,c[2*i].val=b[i].val;
		c[2*i-1].id=2*i-1,c[2*i].id=2*i;
		suma+=a[i].val,sumb+=b[i].val;
	}
	ans=min(suma,sumb);
	sort(c+1,c+2*n+1);
	for(int i=1;i<=2*n;i++){
		sum[i]=sum[i-1]+c[i].val;
		if(c[i].id%2==1)a[(c[i].id+1)/2].id=i;
		else b[c[i].id/2].id=i;
	}
	for(int i=1;i<=n;i++){
		int x;
		if(a[i].id<=n&&b[i].id<=n)x=sum[n];
		else if(a[i].id<=n-1&&b[i].id>n)x=sum[n-1]+c[b[i].id].val;
		else if(b[i].id<=n-1&&a[i].id>n)x=sum[n-1]+c[a[i].id].val;
		else x=sum[n-2]+c[a[i].id].val+c[b[i].id].val;
		ans=min(ans,x);
	}
	printf("%lld\n",ans);
}
```