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 <cstdio>
#include <cmath>
#include <cstring>
#include <iostream>
#include <sstream>

#include <string>
#include <vector>
#include <queue>
#include <list>
#include <set>
#include <map>
#include <unordered_map>
#include <unordered_set>
#include <stack>
#include <deque>
#include <algorithm>
#include <random>
#include <iomanip>
using namespace std;

const long long int mod = 998244353;

int main(int argc, char *argv[]) {
    int n;
    cin >> n;
    vector<int> va, vb, vva, vvb;
    vector<pair<int, int>> vp;
    for (int i = 0 ; i< n ; i++ ) {
        int t;
        cin >> t;
        va.push_back(t);
        vva.push_back(t);
    }    
    for (int i = 0; i < n; i++) {
        int t;
        cin >> t;
        vb.push_back(t);
        vvb.push_back(t);
    }
    sort(vva.begin(), vva.end());
    sort(vvb.begin(), vvb.end());
    unordered_map<int, int> um;
    for (int i = 0; i < n; i++) {
        if (vva[i] > vvb[i]) {
            cout << "No" << endl;
            return 0;
        }
        um[vva[i]] = vvb[i];
    }
    for (int i = 0; i < n; i++) {
        if (um[va[i]] >= vb[i] && va[i] <= vb[i]) {
            cout << "Yes" << endl;
            return 0;
        }
    }
    // cout << "Yes" << endl;
    for (int i = 0; i < n; i++) {
        vp.push_back(make_pair(vb[i], va[i]));
    }
    sort(vp.begin(), vp.end());
    for (int i = 0; i < n; i++) {
        if (vp[i].second >= vva[i] && vp[i].second <= vp[i].first) {
            cout << "Yes" << endl;
            return 0;
        }
    }
    cout << "No" << endl;
}