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: Accepted)

#include <iostream>
#include <vector>
#include <array>
#include <deque>
#include <stack>
#include <queue>
#include <list>
#include <map>
#include <set>
#include <unordered_map>
#include <unordered_set>
#include <algorithm>
#include <numeric>
#include <utility>
#include <cmath>
#include <iomanip>
#include <cassert>
#include <complex>
#include <bitset>
#include <functional>

using namespace std;
using ll=long long int;

int main() {
    int n;
    cin >> n;
    vector<int> a(n);
    for (auto &e: a) cin >> e;
    vector<int> b(n);
    for (auto &e: b) cin >> e;

    vector<pair<int, int>> t(n);
    for (int i = 0; i < n; ++i) {
        t[i].first = b[i];
        t[i].second = a[i];
    }
    sort(t.begin(), t.end());
    for (int i = 0; i < n; ++i) {
        a[i] = t[i].second;
        b[i] = t[i].first;
    }

    vector<int> si(n);
    iota(si.begin(), si.end(), 0);
    sort(si.begin(), si.end(), [&](int l, int r){
        return a[l] < a[r];
    });

    for (int i = 0; i < n; ++i) {
        if (a[si[i]] > b[i]) {
            cout << "No" << endl;
            return 0;
        }
    }

    int len = 0;
    int cur = si[0];
    while (cur != 0) {
        cur = si[cur];
        ++len;
    }

    if (len < n-1) {
        cout << "Yes" << endl;
        return 0;
    }

    for (int i = 0; i < n-1; ++i) {
        if (a[si[i+1]] <= b[i]) {
            cout << "Yes" << endl;
            return 0;
        }
    }

    cout << "No" << endl;

    return 0;
}