#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, n) for (ll i = 0; i < (ll)(n); i++)
ll ans = 0;
const ll INF = 1LL << 60; //MAX 9223372036854775807


int main()
{
ll n ;
cin >> n ;
vector<pair<ll,ll>> v;
ll asum=0,bsum=0;
rep(i,n){
  ll a , b ;
  cin >> a >> b ;
  asum+=a;
  bsum+=b;
  v.push_back({a,i});
  v.push_back({b,i});
}
sort(v.begin(),v.end());
set<ll> s;
rep(i,n){
  ans+=v[i].first;
  s.insert(v[i].second);
}
ll anss=INF;
if(s.size()==n){
  if(v[n].second==v[n-1].second){
    anss=ans;
    ans-=v[n-2].first;
    anss+=v[n+1].first;
    anss-=v[n-1].first;
  }
  else ans-=v[n-1].first;
  ans+=v[n].first;
}
ans=min({ans,asum,bsum,anss});
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;
typedef long long ll;
#define rep(i, n) for (ll i = 0; i < (ll)(n); i++)
ll ans = 0;
const ll INF = 1LL << 60; //MAX 9223372036854775807


int main()
{
ll n ;
cin >> n ;
vector<pair<ll,ll>> v;
ll asum=0,bsum=0;
rep(i,n){
  ll a , b ;
  cin >> a >> b ;
  asum+=a;
  bsum+=b;
  v.push_back({a,i});
  v.push_back({b,i});
}
sort(v.begin(),v.end());
set<ll> s;
rep(i,n){
  ans+=v[i].first;
  s.insert(v[i].second);
}
ll anss=INF;
if(s.size()==n){
  if(v[n].second==v[n-1].second){
    anss=ans;
    ans-=v[n-2].first;
    anss+=v[n+1].first;
    anss-=v[n-1].first;
  }
  else ans-=v[n-1].first;
  ans+=v[n].first;
}
ans=min({ans,asum,bsum,anss});
cout << ans << endl;
return 0;
}

```