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 <bits/stdc++.h>
using namespace std;
#define int long long
#define ll long long
#define P pair<int,int>
#define fi first
#define se second
#define rep(i,n) for(int i=0;i<n;i++)
#define all(v) v.begin(),v.end()
#define pb push_back
template<class T>void chmax(T &a,T b){if(a<b)a=b;}
template<class T>void chmin(T &a,T b){if(a>b)a=b;}
constexpr int INF=1000000000000000000;
constexpr int mod=1000000007;
constexpr double eps=1e-12;
constexpr double pi=3.14159265358979323846264338;
//int dx[]={0,1,0,-1},dy[]={1,0,-1,0};
int gcd(int a,int b){
    if(b==0)return a;
    return gcd(b,a%b);
}
int lcm(int a,int b){
    return a/gcd(a,b)*b;
}
bool prime(int a){
    if(a==1)return false;
    for(int i=2;i*i<=a;i++){
        if(a%i==0)return false;
    }
    return true;
}
int modpow(int a,int b){
    if(b==0)return 1;
    if(b%2)return modpow(a,b-1)*a%mod;
    int memo=modpow(a,b/2);
    return memo*memo%mod;
}
vector<int>kaijo,invkaijo;
void init_fact(int n){
    kaijo.resize(n+1);
    invkaijo.resize(n+1);
    kaijo[0]=1;
    for(int i=1;i<=n;i++){
        kaijo[i]=kaijo[i-1]*i;
        kaijo[i]%=mod;
    }
    rep(i,n+1)invkaijo[i]=modpow(kaijo[i],mod-2);
}
int comb(int a,int b){
    if(a<b)return 0;
    if(a<0||b<0)return 0;
    return kaijo[a]*modpow(kaijo[a-b],mod-2)%mod*modpow(kaijo[b],mod-2)%mod;
}
int inv(int x){
    x=modpow(x,mod-2);
    return x;
}
int n,a[100010],b[100010];
pair<P,int>c[100010];
P d[100010];
bool used[100010];
void dfs(int x){
    used[x]=true;
    if(!used[c[x].se])dfs(c[x].se);
}
signed main(){
    cin.tie(0);ios::sync_with_stdio(false);
    cin>>n;
    rep(i,n)cin>>a[i];
    rep(i,n)cin>>b[i];
    rep(i,n)c[i]={{b[i],a[i]},i};
    sort(a,a+n);
    sort(b,b+n);
    sort(c,c+n);
    rep(i,n){
        if(a[i]>b[i]){
            puts("No");
            return 0;
        }
    }
    rep(i,n-1){
        if(a[i+1]<=b[i]){
            puts("Yes");
            return 0;
        }
    }
    rep(i,n){
        c[i].se=i;
        swap(c[i].fi.fi,c[i].fi.se);
    }
    sort(c,c+n);
    dfs(0);
    rep(i,n){
        if(!used[i]){
            puts("Yes");
            return 0;
        }
    }
    puts("No");
    return 0;
}