#include <bits/stdc++.h>
using namespace std;
using LL=long long;
LL A[100010],B[100010];
int use[100010];
int main(){
  LL ans;
  LL N;scanf("%lld",&N);
  vector<pair<LL,int>>ab;
  for(int i=0;i<N;i++){
    scanf("%lld%lld",&A[i],&B[i]);
    ab.push_back({A[i],i});
    ab.push_back({B[i],i});
  }
  sort(ab.begin(),ab.end());
  int i=0;
  LL sum=0;
  for(i=0;i<N;i++)sum+=ab[i].first;
  for(i=0;i<N;i++)use[ab[i].second]=1;
  for(i=0;i<N;i++)if(use[i]==0)break;
  if(i!=N)ans=sum;
  else{
    if(ab[N].second==ab[N-1].second)
      ans=sum+min(ab[N].first-ab[N-2].first,
                  ab[N+1].first-ab[N-1].first);
    else
      ans=sum+ab[N].first-ab[N-1].first;
  }
  for(i=0,sum=0;i<N;i++)sum+=A[i];
  ans=min(ans,sum);
  
  for(i=0,sum=0;i<N;i++)sum+=B[i];
  ans=min(ans,sum);
  
  printf("%lld\n",ans);
  return 0;
}
### Prompt

Your task is to create 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 <bits/stdc++.h>
using namespace std;
using LL=long long;
LL A[100010],B[100010];
int use[100010];
int main(){
  LL ans;
  LL N;scanf("%lld",&N);
  vector<pair<LL,int>>ab;
  for(int i=0;i<N;i++){
    scanf("%lld%lld",&A[i],&B[i]);
    ab.push_back({A[i],i});
    ab.push_back({B[i],i});
  }
  sort(ab.begin(),ab.end());
  int i=0;
  LL sum=0;
  for(i=0;i<N;i++)sum+=ab[i].first;
  for(i=0;i<N;i++)use[ab[i].second]=1;
  for(i=0;i<N;i++)if(use[i]==0)break;
  if(i!=N)ans=sum;
  else{
    if(ab[N].second==ab[N-1].second)
      ans=sum+min(ab[N].first-ab[N-2].first,
                  ab[N+1].first-ab[N-1].first);
    else
      ans=sum+ab[N].first-ab[N-1].first;
  }
  for(i=0,sum=0;i<N;i++)sum+=A[i];
  ans=min(ans,sum);
  
  for(i=0,sum=0;i<N;i++)sum+=B[i];
  ans=min(ans,sum);
  
  printf("%lld\n",ans);
  return 0;
}
```