#include<bits/stdc++.h>
#define fo(i,a,b) for(int i=a;i<=b;i++)
using namespace std;

typedef long long LL;

const int maxn=1e5+5;

int n,a[2*maxn],p[2*maxn],q[2*maxn];

bool cmp(const int &x,const int &y) {return a[x]<a[y];}

int main()
{
	scanf("%d",&n);
	LL suma=0, sumb=0;
	fo(i,1,n)
	{
		scanf("%d %d",&a[2*i-1],&a[2*i]);
		suma+=a[2*i-1], sumb+=a[2*i];
		p[2*i-1]=2*i-1, p[2*i]=2*i;
	}
	
	LL ans=min(suma,sumb);
	
	sort(p+1,p+1+2*n,cmp);
	fo(i,1,2*n) q[p[i]]=i;
	LL sum=0;
	fo(i,1,n-2) sum+=a[p[i]];
	fo(i,1,n)
	{
		int x=q[2*i-1], y=q[2*i];
		if (x>y) swap(x,y);
		if (y<=n-1) ans=min(ans,sum+a[p[n-1]]+a[p[n]]);
			else if (x<=n-2) ans=min(ans,sum+a[p[n-1]]+a[p[y]]);
				else ans=min(ans,sum+a[p[x]]+a[p[y]]);
	}
	
	printf("%lld\n",ans);
}
### Prompt

Your challenge is to write 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 fo(i,a,b) for(int i=a;i<=b;i++)
using namespace std;

typedef long long LL;

const int maxn=1e5+5;

int n,a[2*maxn],p[2*maxn],q[2*maxn];

bool cmp(const int &x,const int &y) {return a[x]<a[y];}

int main()
{
	scanf("%d",&n);
	LL suma=0, sumb=0;
	fo(i,1,n)
	{
		scanf("%d %d",&a[2*i-1],&a[2*i]);
		suma+=a[2*i-1], sumb+=a[2*i];
		p[2*i-1]=2*i-1, p[2*i]=2*i;
	}
	
	LL ans=min(suma,sumb);
	
	sort(p+1,p+1+2*n,cmp);
	fo(i,1,2*n) q[p[i]]=i;
	LL sum=0;
	fo(i,1,n-2) sum+=a[p[i]];
	fo(i,1,n)
	{
		int x=q[2*i-1], y=q[2*i];
		if (x>y) swap(x,y);
		if (y<=n-1) ans=min(ans,sum+a[p[n-1]]+a[p[n]]);
			else if (x<=n-2) ans=min(ans,sum+a[p[n-1]]+a[p[y]]);
				else ans=min(ans,sum+a[p[x]]+a[p[y]]);
	}
	
	printf("%lld\n",ans);
}
```