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;
typedef long long ll;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
#define all(x) (x).begin(),(x).end()
#define pll pair<ll,ll>
#define pii pair<int,int>
#define rep(i,n) for(int i=0;i<n;i++)
#define sz(x) ((ll)(x).size())
#define pb push_back
#define mp make_pair
#define bit(n) (1LL<<(n))
#define F first
#define S second
int dx[4]={1,0,-1,0};
int dy[4]={0,1,0,-1};
const ll INF = 1LL<<60;
const ll mod = (int)1e9 + 7;

int main() {
    int n;
    cin >> n;
    vector<int> a(n);
    vector<int> b(n);
    rep(i,n)cin >> a[i];
    rep(i,n)cin >> b[i];
    vector<pii> aa(n);
    rep(i,n){
        aa[i]=mp(a[i],b[i]);
    }
    sort(all(aa));
    sort(all(b));
    bool noswa=false;
    rep(i,n){
        if(aa[i].S==b[i])noswa=true;
    }
    if(noswa){
        rep(i,n){
            if(aa[i].F>b[i]){
                cout << "No";
                return 0;
            }
        }
    }
    else{
        rep(i,n-2){
            if(aa[i].F>b[i]){
                cout << "No";
                return 0;
            }
        }
        if(aa[n-2].F>b[n-1] || aa[n-1].F>b[n-2]){
            cout << "No";
            return 0;
        }
    }
    cout << "Yes";

    


}

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