#include<bits/stdc++.h>
#define P pair<int,int>
#define fir first
#define sec second
using namespace std;
typedef long long ll;
const int N=100005;
int n,a[N],b[N],cnt,mp[N];
P p[N*2];
const ll inf=(1ll<<60);
ll ans,now,sum1,sum2;
int main()
{
	scanf("%d",&n);
	for (int i=1;i<=n;i++) {
	   scanf("%d%d",&a[i],&b[i]);
	   p[++cnt]=P(a[i],i);
	   p[++cnt]=P(b[i],i);
	   sum1+=a[i];
	   sum2+=b[i];	
	}
	sort(p+1,p+cnt+1);
	for (int i=1;i<=n;i++)
	{
		now+=p[i].fir;
		mp[p[i].sec]++;
	}
	ans=inf;
	for (int i=1;i<=n;i++)
	{
		if (mp[p[i].sec]==0){
			ans=min(ans,now-p[n].fir-p[n-1].fir+a[i]+b[i]);
		}else if (mp[p[i].sec]==1){
			if (a[i]<b[i]) ans=min(ans,now-((p[n].sec!=i)?p[n].fir:p[n-1].fir)+b[i]);
			else ans=min(ans,now-((p[n].sec!=i)?p[n].fir:p[n-1].fir)+a[i]);
		}else {ans=now;break;}
	}
	printf("%lld\n",min(ans,min(sum1,sum2)));
	return 0;
}
### Prompt

Generate 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 P pair<int,int>
#define fir first
#define sec second
using namespace std;
typedef long long ll;
const int N=100005;
int n,a[N],b[N],cnt,mp[N];
P p[N*2];
const ll inf=(1ll<<60);
ll ans,now,sum1,sum2;
int main()
{
	scanf("%d",&n);
	for (int i=1;i<=n;i++) {
	   scanf("%d%d",&a[i],&b[i]);
	   p[++cnt]=P(a[i],i);
	   p[++cnt]=P(b[i],i);
	   sum1+=a[i];
	   sum2+=b[i];	
	}
	sort(p+1,p+cnt+1);
	for (int i=1;i<=n;i++)
	{
		now+=p[i].fir;
		mp[p[i].sec]++;
	}
	ans=inf;
	for (int i=1;i<=n;i++)
	{
		if (mp[p[i].sec]==0){
			ans=min(ans,now-p[n].fir-p[n-1].fir+a[i]+b[i]);
		}else if (mp[p[i].sec]==1){
			if (a[i]<b[i]) ans=min(ans,now-((p[n].sec!=i)?p[n].fir:p[n-1].fir)+b[i]);
			else ans=min(ans,now-((p[n].sec!=i)?p[n].fir:p[n-1].fir)+a[i]);
		}else {ans=now;break;}
	}
	printf("%lld\n",min(ans,min(sum1,sum2)));
	return 0;
}
```