#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int a[100005],b[100005];
struct info{
	int v,po,t;
	bool operator <(const info b)const{
		return v<b.v;
	}
}c[200005];
int tp[100005],cnt[4];
int main(){
	int n;
	scanf("%d",&n);
	for(int i=1;i<=n;i++)scanf("%d%d",&a[i],&b[i]),c[2*i-1]=(info){a[i],i,1},c[2*i]=(info){b[i],i,2};
	sort(c+1,c+2*n+1);
	for(int i=1;i<=n;i++)tp[c[i].po]|=c[i].t;
	for(int i=1;i<=n;i++)cnt[tp[i]]++;
	long long res=0;
	for(int i=1;i<=n;i++)res+=c[i].v;
	if(cnt[0]||!cnt[1]||!cnt[2])printf("%lld\n",res);
	else if(c[n+1].po!=c[n].po)printf("%lld\n",res+c[n+1].v-c[n].v);
		 else printf("%lld\n",res+min(c[n+1].v-c[n-1].v,c[n+2].v-c[n].v));
}
### 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<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int a[100005],b[100005];
struct info{
	int v,po,t;
	bool operator <(const info b)const{
		return v<b.v;
	}
}c[200005];
int tp[100005],cnt[4];
int main(){
	int n;
	scanf("%d",&n);
	for(int i=1;i<=n;i++)scanf("%d%d",&a[i],&b[i]),c[2*i-1]=(info){a[i],i,1},c[2*i]=(info){b[i],i,2};
	sort(c+1,c+2*n+1);
	for(int i=1;i<=n;i++)tp[c[i].po]|=c[i].t;
	for(int i=1;i<=n;i++)cnt[tp[i]]++;
	long long res=0;
	for(int i=1;i<=n;i++)res+=c[i].v;
	if(cnt[0]||!cnt[1]||!cnt[2])printf("%lld\n",res);
	else if(c[n+1].po!=c[n].po)printf("%lld\n",res+c[n+1].v-c[n].v);
		 else printf("%lld\n",res+min(c[n+1].v-c[n-1].v,c[n+2].v-c[n].v));
}
```