#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
const int maxn=1e5+10;
int n,g[maxn],cnt[4];
pair<int,int> a[maxn<<1];
ll ans;

int main(){
	cin>>n;
	for(int i=1,x,y;i<=n;++i){
		cin>>x>>y;
		a[i*2-1]=make_pair(x,i);
		a[i*2]=make_pair(y,-i);
	}
	sort(a+1,a+2*n+1);
	for(int i=1;i<=n;++i){
		if(a[i].second<0)
			g[-a[i].second]|=1;
		else
			g[a[i].second]|=2;
		ans+=a[i].first;
	}
	for(int i=1;i<=n;++i)
		++cnt[g[i]];
	if(cnt[0]||!cnt[1]||!cnt[2]){
		printf("%lld\n",ans);
		return 0;
	}
	if(abs(a[n].second)==abs(a[n+1].second)&&cnt[a[n].second<0?1:2]>1)
		printf("%lld\n",ans+min(a[n+2].first-a[n].first,a[n+1].first-a[n-1].first));
	else
		printf("%lld\n",ans-a[n].first+a[n+1].first);
	return 0;
}
### Prompt

Please create a solution in cpp 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>
using namespace std;

typedef long long ll;
const int maxn=1e5+10;
int n,g[maxn],cnt[4];
pair<int,int> a[maxn<<1];
ll ans;

int main(){
	cin>>n;
	for(int i=1,x,y;i<=n;++i){
		cin>>x>>y;
		a[i*2-1]=make_pair(x,i);
		a[i*2]=make_pair(y,-i);
	}
	sort(a+1,a+2*n+1);
	for(int i=1;i<=n;++i){
		if(a[i].second<0)
			g[-a[i].second]|=1;
		else
			g[a[i].second]|=2;
		ans+=a[i].first;
	}
	for(int i=1;i<=n;++i)
		++cnt[g[i]];
	if(cnt[0]||!cnt[1]||!cnt[2]){
		printf("%lld\n",ans);
		return 0;
	}
	if(abs(a[n].second)==abs(a[n+1].second)&&cnt[a[n].second<0?1:2]>1)
		printf("%lld\n",ans+min(a[n+2].first-a[n].first,a[n+1].first-a[n-1].first));
	else
		printf("%lld\n",ans-a[n].first+a[n+1].first);
	return 0;
}
```