#include<bits/stdc++.h>
#define int long long
using namespace std;
const int N=1e5+10;
struct node{
	int a,b;
}value[N];
int tota,totb,n,sum[N];
int ans;
int Value[N*2];
signed main(){
	scanf("%lld",&n);
	for(int i=1;i<=n;i++){
		scanf("%lld %lld",&value[i].a,&value[i].b);
		Value[i]=value[i].a,Value[i+n]=value[i].b;
		tota+=value[i].a,totb+=value[i].b;
	}
	ans=min(tota,totb);
	sort(Value+1,Value+1+2*n);
	for(int i=1;i<=n;i++)sum[i]=sum[i-1]+Value[i];
	for(int i=1;i<=n;i++){
		int fi=value[i].a,se=value[i].b;
		if(fi>se)swap(fi,se);
		if(lower_bound(Value+1,Value+1+2*n,fi)-Value<=n-2){
			if(lower_bound(Value+1,Value+1+2*n,se)-Value<=n-1){
				ans=min(ans,sum[n]);
			}
			else ans=min(ans,sum[n-1]+se);	
		}
		else ans=min(ans,sum[n-2]+fi+se);
	}
	printf("%lld\n",ans);
	return 0;
}
### 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<bits/stdc++.h>
#define int long long
using namespace std;
const int N=1e5+10;
struct node{
	int a,b;
}value[N];
int tota,totb,n,sum[N];
int ans;
int Value[N*2];
signed main(){
	scanf("%lld",&n);
	for(int i=1;i<=n;i++){
		scanf("%lld %lld",&value[i].a,&value[i].b);
		Value[i]=value[i].a,Value[i+n]=value[i].b;
		tota+=value[i].a,totb+=value[i].b;
	}
	ans=min(tota,totb);
	sort(Value+1,Value+1+2*n);
	for(int i=1;i<=n;i++)sum[i]=sum[i-1]+Value[i];
	for(int i=1;i<=n;i++){
		int fi=value[i].a,se=value[i].b;
		if(fi>se)swap(fi,se);
		if(lower_bound(Value+1,Value+1+2*n,fi)-Value<=n-2){
			if(lower_bound(Value+1,Value+1+2*n,se)-Value<=n-1){
				ans=min(ans,sum[n]);
			}
			else ans=min(ans,sum[n-1]+se);	
		}
		else ans=min(ans,sum[n-2]+fi+se);
	}
	printf("%lld\n",ans);
	return 0;
}
```