#include<bits/stdc++.h>
#define int long long
#define N 201000
#define min(a,b) (((a)<(b))?(a):(b))
#define max(a,b) (((a)>(b))?(a):(b))
using namespace std;
struct node{
	int val,indx;
	node(){}
	node(int V,int I):val(V),indx(I){}
}A[N],B[N],C[N];
bool operator<(const node a,const node b){
	return a.val<b.val;
}
int sum[N];
signed main(){
	int n;
	scanf("%lld",&n);
	for(int i=1;i<=n;i++)
		scanf("%lld%lld",&A[i].val,&B[i].val);
	int tota,totb;
	tota=totb=0;
	for(int i=1;i<=n;i++){
		C[2*i-1]=(node){A[i].val,2*i-1};
		C[2*i]=(node){B[i].val,2*i};
		tota+=A[i].val,totb+=B[i].val;
	}
	int ans=min(tota,totb);
	sort(C+1,C+2*n+1);
	for(int i=1;i<=2*n;i++){
		sum[i]=sum[i-1]+C[i].val;
		((C[i].indx%2)?A:B)[(C[i].indx+1)>>1].indx=i;
	}
	for(int i=1;i<=n;i++){
		int tmp;
		if(A[i].indx<=n&&B[i].indx<=n)
			tmp=sum[n];
		else if(A[i].indx<=n-1&&B[i].indx>n)
			tmp=sum[n-1]+C[B[i].indx].val;
		else if(A[i].indx>n&B[i].indx<=n-1)
			tmp=sum[n-1]+C[A[i].indx].val;
		else
			tmp=sum[n-2]+C[A[i].indx].val+C[B[i].indx].val;
		ans=min(ans,tmp);
	}
	printf("%lld\n",ans);
	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>
#define int long long
#define N 201000
#define min(a,b) (((a)<(b))?(a):(b))
#define max(a,b) (((a)>(b))?(a):(b))
using namespace std;
struct node{
	int val,indx;
	node(){}
	node(int V,int I):val(V),indx(I){}
}A[N],B[N],C[N];
bool operator<(const node a,const node b){
	return a.val<b.val;
}
int sum[N];
signed main(){
	int n;
	scanf("%lld",&n);
	for(int i=1;i<=n;i++)
		scanf("%lld%lld",&A[i].val,&B[i].val);
	int tota,totb;
	tota=totb=0;
	for(int i=1;i<=n;i++){
		C[2*i-1]=(node){A[i].val,2*i-1};
		C[2*i]=(node){B[i].val,2*i};
		tota+=A[i].val,totb+=B[i].val;
	}
	int ans=min(tota,totb);
	sort(C+1,C+2*n+1);
	for(int i=1;i<=2*n;i++){
		sum[i]=sum[i-1]+C[i].val;
		((C[i].indx%2)?A:B)[(C[i].indx+1)>>1].indx=i;
	}
	for(int i=1;i<=n;i++){
		int tmp;
		if(A[i].indx<=n&&B[i].indx<=n)
			tmp=sum[n];
		else if(A[i].indx<=n-1&&B[i].indx>n)
			tmp=sum[n-1]+C[B[i].indx].val;
		else if(A[i].indx>n&B[i].indx<=n-1)
			tmp=sum[n-1]+C[A[i].indx].val;
		else
			tmp=sum[n-2]+C[A[i].indx].val+C[B[i].indx].val;
		ans=min(ans,tmp);
	}
	printf("%lld\n",ans);
	return 0;
}
```