#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define MN 100005
using namespace std;
typedef long long ll;
int n,tt,a[MN],b[MN],cnt[MN];
ll sumx,sumy,now,ans;
struct node{
	int x,y;
	friend bool operator<(node x,node y){return x.x<y.x;}
}num[MN<<1];
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++)scanf("%d%d",&a[i],&b[i]),num[++tt]=((node){a[i],i}),num[++tt]=((node){b[i],i}),sumx+=a[i],sumy+=b[i];
	sort(num+1,num+2*n+1);
	for(int i=1;i<=n;i++)now+=num[i].x,cnt[num[i].y]++;
	ans=min(sumx,sumy);
	for(int i=1;i<=n;i++)if(cnt[i]>1)return 0*printf("%lld\n",min(ans,now));
	--cnt[num[n].y],++cnt[num[n+1].y];if(cnt[num[n+1].y]>1)return 0*printf("%lld\n",min(ans,now-num[n].x+num[n+1].x));
	printf("%lld\n",min(ans,min(now-num[n-1].x+num[n+1].x,now-num[n].x+num[n+2].x)));
}

### Prompt

In CPP, your task is to solve 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>
#include<cmath>
#define MN 100005
using namespace std;
typedef long long ll;
int n,tt,a[MN],b[MN],cnt[MN];
ll sumx,sumy,now,ans;
struct node{
	int x,y;
	friend bool operator<(node x,node y){return x.x<y.x;}
}num[MN<<1];
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++)scanf("%d%d",&a[i],&b[i]),num[++tt]=((node){a[i],i}),num[++tt]=((node){b[i],i}),sumx+=a[i],sumy+=b[i];
	sort(num+1,num+2*n+1);
	for(int i=1;i<=n;i++)now+=num[i].x,cnt[num[i].y]++;
	ans=min(sumx,sumy);
	for(int i=1;i<=n;i++)if(cnt[i]>1)return 0*printf("%lld\n",min(ans,now));
	--cnt[num[n].y],++cnt[num[n+1].y];if(cnt[num[n+1].y]>1)return 0*printf("%lld\n",min(ans,now-num[n].x+num[n+1].x));
	printf("%lld\n",min(ans,min(now-num[n-1].x+num[n+1].x,now-num[n].x+num[n+2].x)));
}

```