#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll,ll> pll;
int main(){
  ll n;cin>>n;
  ll sa=0,sb=0;
  ll a[n],b[n];
  vector<ll> v;
  for(int i=0;i<n;i++){
    cin>>a[i]>>b[i];
    sa+=a[i];
    sb+=b[i];
    v.push_back(a[i]);
    v.push_back(b[i]);
  }
  sort(v.begin(),v.end());
  ll ss=0;
  for(int i=0;i<n;i++)ss+=v[i];
  ll ans=min(sa,sb);
  for(int i=0;i<n;i++){
    ll sum=a[i]+b[i];
    if(a[i]>b[i])swap(a[i],b[i]);
    ll h=v[n-1];
    ll k=ss;
    if(h>=b[i]){
      k-=a[i]+b[i];
    }
    else if(h>=a[i]){
      if(a[i]>=v[n-3])k-=v[n-1]+v[n-2];
      else k-=v[n-1]+a[i];
    }
    else{
      k-=v[n-1]+v[n-2];
    }
    ans=min(ans,sum+k);
  }
  cout<<ans<<endl;
}

### 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>
using namespace std;
typedef long long ll;
typedef pair<ll,ll> pll;
int main(){
  ll n;cin>>n;
  ll sa=0,sb=0;
  ll a[n],b[n];
  vector<ll> v;
  for(int i=0;i<n;i++){
    cin>>a[i]>>b[i];
    sa+=a[i];
    sb+=b[i];
    v.push_back(a[i]);
    v.push_back(b[i]);
  }
  sort(v.begin(),v.end());
  ll ss=0;
  for(int i=0;i<n;i++)ss+=v[i];
  ll ans=min(sa,sb);
  for(int i=0;i<n;i++){
    ll sum=a[i]+b[i];
    if(a[i]>b[i])swap(a[i],b[i]);
    ll h=v[n-1];
    ll k=ss;
    if(h>=b[i]){
      k-=a[i]+b[i];
    }
    else if(h>=a[i]){
      if(a[i]>=v[n-3])k-=v[n-1]+v[n-2];
      else k-=v[n-1]+a[i];
    }
    else{
      k-=v[n-1]+v[n-2];
    }
    ans=min(ans,sum+k);
  }
  cout<<ans<<endl;
}

```