Problem:
Score : 600 points

Problem Statement

Given are two integer sequences of N elements each: A_1,...,A_N and B_1,...,B_N.
Determine if it is possible to do the following operation at most N-2 times (possibly zero) so that, for every integer i from 1 to N, A_i \leq B_i holds:

Choose two distinct integers x and y between 1 and N (inclusive), and swap the values of A_x and A_y.

Constraints

2 \leq N \leq 10^5

1 \leq A_i,B_i \leq 10^9

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N
B_1 B_2 ... B_N

Output

If the objective is achievable, print Yes; if it is not, print No.

Sample Input 1

3
1 3 2
1 2 3

Sample Output 1

Yes

We should swap the values of A_2 and A_3.

Sample Input 2

3
1 2 3
2 2 2

Sample Output 2

No

Sample Input 3

6
3 1 2 6 3 4
2 2 8 3 4 3

Sample Output 3

Yes

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

#include <iostream>
#include <vector>
#include <queue>
#include <utility>

using namespace std;

int main(){
    int n;
    cin >> n;
    vector<pair<int, int>> p(n);
    for(int i = 0; i < n; i++){
        cin >> p[i].second;
        p[i].second *= -1;
    }
    for(int i = 0; i < n; i++){
        cin >> p[i].first;
    }
    priority_queue<pair<int, int>> success, fail;
    for(auto&& i : p){
        if(i.first == -i.second) continue;
        if(i.first > -i.second){
            success.push(i);
        }else{
            fail.push(make_pair(-i.second, i.first));
        }
    }
    int cnt = 0;
    while(!success.empty() && cnt < n - 2 && !fail.empty()){
        pair<int, int> s;
        do{
            s = success.top();
            success.pop();
        }while(!success.empty() && -s.second >= fail.top().first);
        
        if(-s.second >= fail.top().first || s.first < fail.top().first){
            cout << "No\n";
            return 0;
        }
        auto a = fail.top();
        fail.pop();
        a.first = -s.second;
        if(a.first < a.second){
            success.push(make_pair(a.second, -a.first));
        }else if(a.second > a.first){
            fail.push(a);
        }
        cnt++;
    }
    if(fail.empty()){
        cout << "Yes\n";
    }else{
        cout << "No\n";
    }
}