#include <bits/stdc++.h>
#define MOD 1000000007LL
using namespace std;
typedef long long ll;
typedef pair<ll,ll> P;

int n;
ll a[100005];
ll b[100005];
P p[200005];
int cnt[100005];


int main(void){
	scanf("%d",&n);
	ll ans=0;
	for(int i=0;i<n;i++){
		scanf("%lld%lld",&a[i],&b[i]);
		ans+=b[i];
		p[i]=P(a[i],i);
		p[i+n]=P(b[i],i);
	}
	ll val=0;
	for(int i=0;i<n;i++){
		val+=a[i];
	}
	ans=min(ans,val);
	sort(p,p+2*n);
	bool flag=false;
	for(int i=0;i<n;i++){
		cnt[p[i].second]++;
		if(cnt[p[i].second]>=2){
			flag=true;
		}
	}
	ll sum=0;
	for(int i=0;i<n;i++){
		sum+=p[i].first;
	}
	if(flag){	
		printf("%lld\n",sum);
	}else{
		for(int i=0;i<n;i++){
			if(p[n].second==i){
				ans=min(ans,sum-min(a[i],b[i])+p[n+1].first);
			}else{
				ans=min(ans,sum-min(a[i],b[i])+p[n].first);
			}
		}
		printf("%lld\n",ans);
	}
	return 0;
} 
### Prompt

Develop a solution in CPP to the problem described below:

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 MOD 1000000007LL
using namespace std;
typedef long long ll;
typedef pair<ll,ll> P;

int n;
ll a[100005];
ll b[100005];
P p[200005];
int cnt[100005];


int main(void){
	scanf("%d",&n);
	ll ans=0;
	for(int i=0;i<n;i++){
		scanf("%lld%lld",&a[i],&b[i]);
		ans+=b[i];
		p[i]=P(a[i],i);
		p[i+n]=P(b[i],i);
	}
	ll val=0;
	for(int i=0;i<n;i++){
		val+=a[i];
	}
	ans=min(ans,val);
	sort(p,p+2*n);
	bool flag=false;
	for(int i=0;i<n;i++){
		cnt[p[i].second]++;
		if(cnt[p[i].second]>=2){
			flag=true;
		}
	}
	ll sum=0;
	for(int i=0;i<n;i++){
		sum+=p[i].first;
	}
	if(flag){	
		printf("%lld\n",sum);
	}else{
		for(int i=0;i<n;i++){
			if(p[n].second==i){
				ans=min(ans,sum-min(a[i],b[i])+p[n+1].first);
			}else{
				ans=min(ans,sum-min(a[i],b[i])+p[n].first);
			}
		}
		printf("%lld\n",ans);
	}
	return 0;
} 
```