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 "bits/stdc++.h"

using namespace std;
//vector<long long> factors;
#define ll long long

#define pii pair<int,int>
#define pll pair<ll,ll>
#define _ <<'\n'
#define __ << ' '
#define all(x) (x).begin(), (x).end()
#define gcd __gcd
int IT_MAX = 1 << 17;
int MOD = 998244353;
const int INF = 0x3f3f3f3f;
const ll LL_INF = 0x0f3f3f3f3f3f3f3f;
const double PI = acos(-1);
const double ERR = 1e-10;
#define szz(x) (int)(x).size()
#define IOS ios_base::sync_with_stdio(false); cout.tie(0); cin.tie(0);
#define flush fflush(stdout)

ll mod(ll a, ll m) { return (a + (abs(a) / m) * m + m) % m; }

//****************************don't use int!!!!!!!****************************//
//***************************code starts here!!!!!****************************//


int main() {
    IOS

    ll n;
    cin >> n;
    vector<ll> a(n);
    vector<ll> b(n);
    for(int i = 0;i < n; ++i){
        cin >> a[i];
    }

    for(int i = 0;i < n; ++i){
        cin >> b[i];
    }
    sort(all(a));
    sort(all(b));
    for(int i = 0;i < n ;++i){
        if(a[i] > b[i]){
            cout << "No";
            return 0;
        }
    }
    cout << "yes";

    return 0;
}