#include<bits/stdc++.h>
using namespace std;
#define N 100005
#define pb push_back
#define pii pair<int,int>
#define mp make_pair
#define LL long long
#define F first
#define S second
 
int n;
LL ans;
LL sum1,sum2;
bool ok[N];
vector<pii> v;
 
int main()
{
	
	cin>>n;
	for(int i=1;i<=n;i++)
	{
		int x,y;
		cin>>x>>y;
		sum1+=x; sum2+=y;
		v.pb(mp(x,i)); v.pb(mp(y,i));
	}
	sort(v.begin(),v.end());
	for(int i=0;i<n;i++)
		ans+=v[i].F;
	for(int i=0;i<n;i++)
	{
		if(ok[v[i].S]) return cout<<ans<<'\n',0;
		ok[v[i].S]=1;
	}
	cout<<min(min(sum1,sum2),ans-v[n-1].F+((v[n-1].S!=v[n].S)?v[n].F:min(v[n+1].F,v[n-1].F-v[n-2].F+v[n].F)))<<'\n';
	
	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>
using namespace std;
#define N 100005
#define pb push_back
#define pii pair<int,int>
#define mp make_pair
#define LL long long
#define F first
#define S second
 
int n;
LL ans;
LL sum1,sum2;
bool ok[N];
vector<pii> v;
 
int main()
{
	
	cin>>n;
	for(int i=1;i<=n;i++)
	{
		int x,y;
		cin>>x>>y;
		sum1+=x; sum2+=y;
		v.pb(mp(x,i)); v.pb(mp(y,i));
	}
	sort(v.begin(),v.end());
	for(int i=0;i<n;i++)
		ans+=v[i].F;
	for(int i=0;i<n;i++)
	{
		if(ok[v[i].S]) return cout<<ans<<'\n',0;
		ok[v[i].S]=1;
	}
	cout<<min(min(sum1,sum2),ans-v[n-1].F+((v[n-1].S!=v[n].S)?v[n].F:min(v[n+1].F,v[n-1].F-v[n-2].F+v[n].F)))<<'\n';
	
	return 0;
}
```