#include<cstdio>
#include<algorithm>
using namespace std;
const int N=2e5+5;
int n,id[N];long long sum,ans;
pair<int,int>a[N];
int main(){
	scanf("%d",&n);
	for(int i=1,x,y;i<=n;++i){
		scanf("%d%d",&x,&y);
		ans+=x;sum+=y;
		a[i]={x,i};a[i+n]={y,i+n};
	}
	ans=min(ans,sum);sum=0;
	sort(a+1,a+n+n+1);
	for(int i=1;i<=n+n;++i)id[a[i].second]=i;
	for(int i=1;i<=n;++i)sum+=a[i].first;
	for(int i=1;i<=n;++i){
		int x=id[i],y=id[i+n];
		if(x>y)swap(x,y);
		if(y<=n)ans=min(ans,sum);
		else if(x<n)ans=min(ans,sum-a[n].first+a[y].first);
		else ans=min(ans,sum-a[n].first-a[n-1].first+a[x].first+a[y].first);
	}
	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<cstdio>
#include<algorithm>
using namespace std;
const int N=2e5+5;
int n,id[N];long long sum,ans;
pair<int,int>a[N];
int main(){
	scanf("%d",&n);
	for(int i=1,x,y;i<=n;++i){
		scanf("%d%d",&x,&y);
		ans+=x;sum+=y;
		a[i]={x,i};a[i+n]={y,i+n};
	}
	ans=min(ans,sum);sum=0;
	sort(a+1,a+n+n+1);
	for(int i=1;i<=n+n;++i)id[a[i].second]=i;
	for(int i=1;i<=n;++i)sum+=a[i].first;
	for(int i=1;i<=n;++i){
		int x=id[i],y=id[i+n];
		if(x>y)swap(x,y);
		if(y<=n)ans=min(ans,sum);
		else if(x<n)ans=min(ans,sum-a[n].first+a[y].first);
		else ans=min(ans,sum-a[n].first-a[n-1].first+a[x].first+a[y].first);
	}
	printf("%lld\n",ans);
	return 0;
}

```