#include<bits/stdc++.h>
using namespace std;
#define LL long long
#define N 100011
struct PEO{
	LL val;int bel;
}p[N<<1];
int n,ba[N],bb[N];
LL a[N],b[N],ans,sum[N<<1];
bool cmp(PEO a,PEO b){
	return a.val<b.val;
}
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++)scanf("%lld%lld",&a[i],&b[i]);
	LL tmp=0;
	for(int i=1;i<=n;i++)ans=ans+a[i],tmp=tmp+b[i];
	ans=min(ans,tmp);
	for(int i=1;i<=n;i++)p[i].val=a[i],p[i].bel=i;
	for(int i=1;i<=n;i++)p[n+i].val=b[i],p[n+i].bel=i+n;
	sort(p+1,p+2*n+1,cmp);
	for(int i=1;i<=2*n;i++){
		if(p[i].bel<=n) ba[p[i].bel]=i;
		else bb[p[i].bel-n]=i;
		sum[i]=sum[i-1]+p[i].val;
	}
	for(int i=1;i<=n;i++){
		if(bb[i]<=n&&ba[i]<=n) ans=min(ans,sum[n]);
		else if(ba[i]<=n){
			if(ba[i]==n){
				ans=min(ans,sum[n]-p[n-1].val+b[i]);
			}else
				ans=min(ans,sum[n-1]+b[i]);
		}else
		if(bb[i]<=n){
			if(bb[i]==n){
				ans=min(ans,sum[n]-p[n-1].val+a[i]);
			}else{
				ans=min(ans,sum[n-1]+a[i]);
			}
		}
	}
	printf("%lld\n",ans);
} 
### Prompt

Construct a Cpp code solution to the problem outlined:

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
#define N 100011
struct PEO{
	LL val;int bel;
}p[N<<1];
int n,ba[N],bb[N];
LL a[N],b[N],ans,sum[N<<1];
bool cmp(PEO a,PEO b){
	return a.val<b.val;
}
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++)scanf("%lld%lld",&a[i],&b[i]);
	LL tmp=0;
	for(int i=1;i<=n;i++)ans=ans+a[i],tmp=tmp+b[i];
	ans=min(ans,tmp);
	for(int i=1;i<=n;i++)p[i].val=a[i],p[i].bel=i;
	for(int i=1;i<=n;i++)p[n+i].val=b[i],p[n+i].bel=i+n;
	sort(p+1,p+2*n+1,cmp);
	for(int i=1;i<=2*n;i++){
		if(p[i].bel<=n) ba[p[i].bel]=i;
		else bb[p[i].bel-n]=i;
		sum[i]=sum[i-1]+p[i].val;
	}
	for(int i=1;i<=n;i++){
		if(bb[i]<=n&&ba[i]<=n) ans=min(ans,sum[n]);
		else if(ba[i]<=n){
			if(ba[i]==n){
				ans=min(ans,sum[n]-p[n-1].val+b[i]);
			}else
				ans=min(ans,sum[n-1]+b[i]);
		}else
		if(bb[i]<=n){
			if(bb[i]==n){
				ans=min(ans,sum[n]-p[n-1].val+a[i]);
			}else{
				ans=min(ans,sum[n-1]+a[i]);
			}
		}
	}
	printf("%lld\n",ans);
} 
```