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>
#define FOR(i, a, n) for(ll i = (ll)a; i < (ll)n; i++)
#define rep(i, n) FOR(i, 0, n)
#define ALL(x) (x).begin(), (x).end()
using namespace std;
typedef long long ll;

constexpr ll mod = 1e9 + 7;
constexpr ll inf = 1LL << 60;

template <typename T> constexpr bool chmax(T &a, const T b) {
    if(a >= b) return false;
    a = b;
    return true;
}
template <typename T> constexpr bool chmin(T &a, const T b) {
    if(a <= b) return false;
    a = b;
    return true;
}

/*-------------------------------------------*/

int main() {
    cin.tie(0);
    ios::sync_with_stdio(0);

    int n;
    cin >> n;
    vector<pair<int, int>> v;
    rep(i, n) {
        int a, b;
        cin >> a >> b;
        v.push_back({a, i});
        v.push_back({b, i + n});
    }
    sort(ALL(v));

    bool a = true, b = true, c[n] = {}, d = false;
    ll sum = 0;
    rep(i, n) {
        sum += v[i].first;
        (v[i].second < n ? a : b) = false;
        (c[v[i].second % n] ? d : c[v[i].second % n]) = true;
    }
    if(d || a || b)
        cout << sum << endl;
    else if(abs(v[n - 1].second - v[n].second) != n)
        cout << sum - v[n - 1].first + v[n].first << endl;
    else
        cout << min(sum - v[n - 1].first + v[n + 1].first,
                    sum - v[n - 2].second + v[n].second)
             << endl;

    return 0;
}