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<stdio.h>
#include<string>
#include<string.h>
#include<utility>
#include<iostream>
#include<queue>
#include<math.h>
#include<algorithm>
#include<vector>
#include<map>
#include<stack>
#include<bitset>
#include <cmath>
#include <limits>
#include <numeric>
#include <cmath>
#include <limits>

using namespace std;
typedef long long ll;
#define REP(i,n) for(int i=0;i<(int)(n);i++)

int main(){
    int n;
    ll a[200005]={};
    ll b[200005]={};
    ll aa[200005]={};
    ll bb[200005]={};
    scanf("%d",&n);
    REP(i,n)scanf("%lld",&a[i]);
    REP(i,n)scanf("%lld",&b[i]);
    REP(i,n){
        aa[i]=a[i];
        bb[i]=b[i];
    }
    sort(a,a+n,std::greater<ll>());
    sort(b,b+n,std::greater<ll>());
    int flag=0;
    REP(i,n){
        if(a[i]>b[i]){
            printf("No\n");
            return 0;
        }
        else{
            if(a[i]<=b[i+1]){
                flag=1;
            }
        }
    }
    if(flag==1){
        printf("Yes\n");
        return 0;
    }
    std::map<ll,ll> ma;
    REP(i,n)ma[a[i]]=i;
    ll aaa[200005]={};
    REP(i,n)aaa[i]=ma[aa[i]];
    
    std::map<ll,ll> mb;
    REP(i,n)mb[b[i]]=i;
    ll bbb[200005]={};
    REP(i,n)bbb[i]=mb[bb[i]];
    
    std::map<ll,ll> maaa;
    REP(i,n)maaa[aaa[i]]=i;
    int now=bbb[0];
    REP(i,n){
        now = bbb[maaa[now]];
        if(i!=n-1 && now==bbb[0]){
            printf("No\n");
            return 0;
        }
    }
    printf("Yes\n");
    return 0;
}