#include<bits/stdc++.h>
#define rep(i,x,y) for (int i=(x);i<=(y);i++)
#define ll long long
#define pii pair<int,int>
#define fi first
#define se second
using namespace std;
const int N=2e5+5;
int n,x,y,vis[N],flag; pii a[N]; ll s1,s2,s3;
int main(){
	scanf("%d",&n);
	rep (i,1,n){
		scanf("%d%d",&x,&y);
		a[2*i-1]=pii(x,i),a[2*i]=pii(y,i);
		s1+=x,s2+=y;
	}
	sort(a+1,a+1+2*n);
	rep (i,1,n){
		s3+=a[i].fi;
		if (vis[a[i].se]) flag=1;
		vis[a[i].se]=1;
	}
	if (!flag)
		if (a[n].se==a[n+1].se) s3+=min(a[n+2].fi-a[n].fi,a[n+1].fi-a[n-1].fi);
		else s3+=a[n+1].fi-a[n].fi;
	printf("%lld\n",min(s1,min(s2,s3)));
	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 rep(i,x,y) for (int i=(x);i<=(y);i++)
#define ll long long
#define pii pair<int,int>
#define fi first
#define se second
using namespace std;
const int N=2e5+5;
int n,x,y,vis[N],flag; pii a[N]; ll s1,s2,s3;
int main(){
	scanf("%d",&n);
	rep (i,1,n){
		scanf("%d%d",&x,&y);
		a[2*i-1]=pii(x,i),a[2*i]=pii(y,i);
		s1+=x,s2+=y;
	}
	sort(a+1,a+1+2*n);
	rep (i,1,n){
		s3+=a[i].fi;
		if (vis[a[i].se]) flag=1;
		vis[a[i].se]=1;
	}
	if (!flag)
		if (a[n].se==a[n+1].se) s3+=min(a[n+2].fi-a[n].fi,a[n+1].fi-a[n-1].fi);
		else s3+=a[n+1].fi-a[n].fi;
	printf("%lld\n",min(s1,min(s2,s3)));
	return 0;
}
```