#include<bits/stdc++.h>
using namespace std;
using ll=long long;
using P=pair<ll,ll>;
ll N;
int main(){
  cin>>N;ll a=0;ll b=0;vector<P>A;
  for(int i=0;i<N;i++){
  ll x,y;cin>>x>>y;
    a+=x;b+=y;P p(x,i);P q(y,i);
    A.push_back(p);
    A.push_back(q);
  }sort(A.begin(),A.end());
  set<ll>S;ll ans=0;
  for(int i=0;i<N;i++){
  ans+=A[i].first;
    S.insert(A[i].second);
  }ll M=S.size();
  if(M==N){
  if(A[N].second!=A[N-1].second)
    ans+=(A[N].first-A[N-1].first);
    else{
      ans+=min(A[N].first-A[N-2].first,A[N+1].first-A[N-1].first);
    }
   }a=min(a,b);ans=min(a,ans);
  cout<<ans<<endl;
  return 0;
}

### Prompt

Create a solution in cpp for 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;
using P=pair<ll,ll>;
ll N;
int main(){
  cin>>N;ll a=0;ll b=0;vector<P>A;
  for(int i=0;i<N;i++){
  ll x,y;cin>>x>>y;
    a+=x;b+=y;P p(x,i);P q(y,i);
    A.push_back(p);
    A.push_back(q);
  }sort(A.begin(),A.end());
  set<ll>S;ll ans=0;
  for(int i=0;i<N;i++){
  ans+=A[i].first;
    S.insert(A[i].second);
  }ll M=S.size();
  if(M==N){
  if(A[N].second!=A[N-1].second)
    ans+=(A[N].first-A[N-1].first);
    else{
      ans+=min(A[N].first-A[N-2].first,A[N+1].first-A[N-1].first);
    }
   }a=min(a,b);ans=min(a,ans);
  cout<<ans<<endl;
  return 0;
}

```