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: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define int long long

vector<pair<int,pair<bool, int> > > v;
int n;
int something[100005];
int l[100005];
int r[100005];
int sumleft = 0;
int sumright = 0;

main(){
scanf("%lld",&n);
for (int x = 0; x<n; x++){
    int a,b;
    scanf("%lld%lld",&a,&b);
    v.push_back({a,{false,x}});
    v.push_back({b,{true,x}});
    sumleft+=a;
    sumright+=b;
    l[x] = a;
    r[x] = b;

}
sort(v.begin(),v.end());
int ans = 0;
bool possible = false;
for (int x = 0; x<n; x++){
    something[v[x].second.second]++;
    if (something[v[x].second.second]>1){
        possible = true;
    }
    ans += v[x].first;


}


if (possible){
    printf("%lld",ans);
    return 0;
}
int ans2 = min(sumleft,sumright);
if (ans == ans2){
    printf("%lld",ans);
    return 0;
}
for (int x = 0; x<n; x++){
    if (x==v[n-1].second.second){
        ans2 = min(ans2,ans-v[n-2].first+max(l[x],r[x]));
    }
    else{
        ans2 = min(ans2,ans-v[n-1].first + max(l[x],r[x]));
    }
}


printf("%lld",ans2);

}