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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#include <random>

using namespace std;

typedef unsigned long long ull;
typedef long long ll;
typedef double ld;
//#define ll __int128
//#define int ll
//#define int ll
//#define char ll
#define double ld
typedef vector<char> vc;
typedef vector<vc> vvc;
typedef vector<vvc> vvvc;
typedef pair<int, int> pii;
typedef pair<pii, pii> piii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<pii> vpi;
typedef vector<vi> vvi;
typedef vector<vvi> vvvi;
typedef vector<short> vs;
typedef vector<vs> vvs;
typedef vector<vvs> vvvs;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<vvl> vvvl;
typedef vector<ld> vld;
typedef vector<vld> vvld;
typedef vector<vvld> vvvld;
typedef vector<string> vst;
typedef vector<vst> vvst;
typedef pair<ld, ld> pld;

#define inmin(a, b) a = min(a, (b))
#define inmax(a, b) a = max(a, (b))
#define ALL(a) a.begin(),a.end()
#define RALL(a) a.rbegin(),a.rend()
#define sqr(x) ((x) * (x))
#define fori(i, n) for(int i = 0; i < int(n); ++i)
#define SZ(a) ((int)((a).size()))
#define triple(T) tuple<T, T, T>
#define quad(T) tuple<T, T, T, T>
#define watch(x) cerr << (#x) << " = " << (x) << endl;

#ifdef MAX_HOME
#define cerr cout
#else
#define cerr if (false) cerr
#endif

const double PI = 2 * acos(0.0);
#define rand shittttty_shit
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
mt19937_64 rng_64(chrono::steady_clock::now().time_since_epoch().count());

const string DIGITS = "0123456789";
const string ALPH = "abcdefghijklmnopqrstuvwxyz";


template<class T0, class T1>
inline ostream &operator<<(ostream &out, pair<T0, T1> &a) {
    return out << "{" << a.first << ", " << a.second << "}";
}

template<class T0, class T1>
inline istream &operator>>(istream &in, pair<T0, T1> &a) {
    return in >> a.first >> a.second;
}

template<class T0, class T1, class T2>
inline ostream &operator<<(ostream &out, tuple<T0, T1, T2> &a) {
    return out << "{" << get<0>(a) << ", " << get<1>(a) << ", " << get<2>(a) << "}";
}

template<class T0, class T1, class T2, class T3>
inline ostream &operator<<(ostream &out, tuple<T0, T1, T2, T3> &a) {
    return out << "{" << get<0>(a) << ", " << get<1>(a) << ", " << get<2>(a) << ", " << get<3>(a) << "}";
}

template<class T>
inline ostream &operator<<(ostream &out, vector<T> &a) {
    out << "[";
    fori (i, a.size())out << a[i] << vector<string>{", ", "]  "}[i + 1 == a.size()];
    return out;
}


void smain();


signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
#ifdef MAX_HOME
    freopen("input.txt", "r", stdin);
    clock_t start = clock();
#endif
    cout << setprecision(12) << fixed;
    smain();
#ifdef MAX_HOME
    cout << "\n\nTOTAL EXECUTION TIME: " << float(clock() - start) / CLOCKS_PER_SEC << endl;
#endif
    return 0;
}

const int N = 2e5 + 10;

pii a[N], b[N];

void bad() {
    cout << "No\n";
    exit(0);
}

void smain() {
    int n;
    cin >> n;
    fori (i, n) {
        cin >> a[i].first;
        a[i].second = i;
    }
    fori (i, n) {
        cin >> b[i].first;
        b[i].second = i;
    }
    sort(a, a + n);
    sort(b, b + n);
    fori (i, n) {
        if (a[i].first > b[i].first)
            bad();
    }
    vi p(n);
    fori (i, n) {
        if (a[i].first <= b[i + 1].first) {
            cout << "Yes\n";
            return;
        }
        p[a[i].second] = b[i].second;
    }
    int cur = 0;
    int ans = 1;
    while (p[cur]) {
        cur = p[cur];
        ans++;
    }
    cout << (ans == n ? "No\n" : "Yes\n");
}