#include<cstdio>
#include<algorithm>
#define rep(i,l,r) for (int i=(l); i<=(r); i++)
typedef long long ll;
using namespace std;

const int N=200010;
struct P{ int x,y; }p[N];
bool cmp(const P &a,const P &b){ return a.x<b.x; }
int n,x,y,u[N];
ll ans;

int main(){
	scanf("%d",&n);
	rep(i,1,n) scanf("%d%d",&p[i].x,&p[i+n].x),p[i].y=i,p[i+n].y=-i;
	sort(p+1,p+n+n+1,cmp);
	rep(i,1,n) ans+=p[i].x,x|=u[abs(p[i].y)]++,y|=(1ll*p[i].y*p[i-1].y<0);
	printf("%lld\n",(x||!y) ? ans : (p[n].y==-p[n+1].y ? ans+min(p[n+2].x-p[n].x,p[n+1].x-p[n-1].x) : ans-p[n].x+p[n+1].x));
	return 0;
}
### 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<cstdio>
#include<algorithm>
#define rep(i,l,r) for (int i=(l); i<=(r); i++)
typedef long long ll;
using namespace std;

const int N=200010;
struct P{ int x,y; }p[N];
bool cmp(const P &a,const P &b){ return a.x<b.x; }
int n,x,y,u[N];
ll ans;

int main(){
	scanf("%d",&n);
	rep(i,1,n) scanf("%d%d",&p[i].x,&p[i+n].x),p[i].y=i,p[i+n].y=-i;
	sort(p+1,p+n+n+1,cmp);
	rep(i,1,n) ans+=p[i].x,x|=u[abs(p[i].y)]++,y|=(1ll*p[i].y*p[i-1].y<0);
	printf("%lld\n",(x||!y) ? ans : (p[n].y==-p[n+1].y ? ans+min(p[n+2].x-p[n].x,p[n+1].x-p[n-1].x) : ans-p[n].x+p[n+1].x));
	return 0;
}
```