#include<bits/stdc++.h>
using namespace std;
const int N=1e5+5;
int n,a[N],b[N],p[N],q[N],i,mn;
pair<int,int*>c[N*2];
long long s[N*2],ans=1ll<<60,t,sa;
int main(){
	scanf("%d",&n);
	for(i=1;i<=n;++i)scanf("%d%d",a+i,b+i),c[i*2-1]=make_pair(a[i],p+i),c[i*2]=make_pair(b[i],q+i),sa+=a[i];
	sort(c+1,c+n*2+1);for(i=1;i<=n*2;++i)*c[i].second=i,s[i]=s[i-1]+c[i].first;
	ans=min(ans,min(sa,s[n*2]-sa));
	for(i=1;i<=n;++i){
		int x=p[i],y=q[i];if(x>y)swap(x,y);
		if(y<=n)ans=min(ans,s[n]);else if(x<=n-1)ans=min(ans,s[n-1]+c[y].first);
			else ans=min(ans,s[n-2]+a[i]+b[i]);
	}
	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<bits/stdc++.h>
using namespace std;
const int N=1e5+5;
int n,a[N],b[N],p[N],q[N],i,mn;
pair<int,int*>c[N*2];
long long s[N*2],ans=1ll<<60,t,sa;
int main(){
	scanf("%d",&n);
	for(i=1;i<=n;++i)scanf("%d%d",a+i,b+i),c[i*2-1]=make_pair(a[i],p+i),c[i*2]=make_pair(b[i],q+i),sa+=a[i];
	sort(c+1,c+n*2+1);for(i=1;i<=n*2;++i)*c[i].second=i,s[i]=s[i-1]+c[i].first;
	ans=min(ans,min(sa,s[n*2]-sa));
	for(i=1;i<=n;++i){
		int x=p[i],y=q[i];if(x>y)swap(x,y);
		if(y<=n)ans=min(ans,s[n]);else if(x<=n-1)ans=min(ans,s[n-1]+c[y].first);
			else ans=min(ans,s[n-2]+a[i]+b[i]);
	}
	printf("%lld\n",ans);
}

```