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 Rust (1.15.1) (judge verdict: Wrong Answer)

use std::io::Read;

fn main() {
    let mut buf = String::new();
    std::io::stdin().read_to_string(&mut buf).ok();
    let mut it = buf.split_whitespace();
    let n = it.next().unwrap().parse::<usize>().unwrap();
    let mut a = Vec::new();
    for i in 0..n {
        for j in 0..2 {
            let v = it.next().unwrap().parse::<i64>().unwrap();
            a.push((v, i, j));
        }
    }
    a.sort();
    let mut res = 0;
    let mut ok = false;
    let mut cnt_in = 0;
    let mut used = vec![0; n];
    for i in 0..n {
        res += a[i].0;
        used[a[i].1] += 1;
        if used[a[i].1] == 2 { ok = true; }
        if a[i].2 == 0 { cnt_in += 1; }
    }
    if cnt_in != n && !ok {
        let mut dif = 1_000_000_000_000_000_000;
        for i in 0..n {
            dif = std::cmp::min(dif, a[if a[i].2 != a[n].2 { n } else { n+1 }].0 - a[i].0);
        }
        res += dif;
    }
    println!("{}", res);
}