Problem:
Score : 700 points

Problem Statement

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.

Sample Input 1

3
1 5
4 2
6 3

Sample Output 1

7

Consider the cycle 1→3→2→1. The weights of those edges are {\rm min}(A_1,B_3)=1, {\rm min}(A_3,B_2)=2 and {\rm min}(A_2,B_1)=4, for a total of 7.
As there is no cycle with a total weight of less than 7, the answer is 7.

Sample Input 2

4
1 5
2 6
3 7
4 8

Sample Output 2

10

Sample Input 3

6
19 92
64 64
78 48
57 33
73 6
95 73

Sample Output 3

227

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

        #include <bits/stdc++.h>
        using namespace std;
   #define int long long
   #define PP pair<int,pair<int,int>>
const int MAX = 510000;
const int MOD =  1000000007;
const int INV2 = (MOD+1)/2;

     signed main(){
       int n,ans=0;
       cin>>n;
       bool used[n]={},t=false;
       pair<pair<int,int>,int> p[2*n];
       int a,b,mina=MOD,minb=MOD,maxa=-1,maxb=-1,suma=0,sumb=0;
       for(int i=0;i<n;i++){
         cin>>a>>b;
         mina=min(a,mina);
         maxa=max(a,maxa);
         suma+=a;
         minb=min(b,minb);
         maxb=max(b,maxb);
         sumb+=b;
         p[2*i]=make_pair(make_pair(a,b),i);
         p[2*i+1]=make_pair(make_pair(b,a),i);
       }
       if(maxa<=minb){
         cout<<suma<<endl;
         return 0;
       }
       if(maxb<=mina){
         cout<<sumb<<endl;
         return 0;
       }
       sort(p,p+2*n);
       for(int i=0;i<n;i++){
         ans+=p[i].first.first;
         if(used[p[i].second])t=true;
         else used[p[i].second]=true;
       }
       if(t)cout<<ans<<endl;
       else {
         if(p[n-1].second==p[n].second)cout<<ans-p[n-1].first.first+p[n+1].first.first<<endl;
         else cout<<ans-p[n-1].first.first+p[n].first.first<<endl;
       } 

                  
}