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 <iostream>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <map>
#include <set>
#include <algorithm>
#include <queue>
#include <stack>
#include <functional>

using namespace std;
typedef long long ll;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef pair<ll,ll> P;
typedef vector<P> vpl;
#define rep(i,n) for(ll i=0; i<(n); i++)
#define REP(i,a,b) for(ll i=(a); i<(b); i++)
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
const int inf = 1<<30;
const ll linf = 1LL<<62;
const int mod = 1e9+7;
const int MAX = 510000;
const int V = 100005;
ll dy[8] = {1,0,-1,0,1,-1,1,-1};
ll dx[8] = {0,1,0,-1,1,-1,-1,1};
const double pi = acos(-1);
template<typename T1,typename T2> inline void chmin(T1 &a,T2 b){if(a>b) a=b;}
template<typename T1,typename T2> inline void chmax(T1 &a,T2 b){if(a<b) a=b;}
template<typename T1,typename T2> inline void print2(T1 a, T2 b){cout << a << " " << b << endl;}
#define INT_MAX 1e9+7

struct UnionFind{
    vl p;
    vl rank;
    vl cnt;

    UnionFind(ll n){
        rank.resize(n,0);
        p.resize(n,0);
        cnt.resize(n,0);
        rep(i,n){
            p[i] = i;
            rank[i] = 0;
            cnt[i] = 1;
        }
    }

    ll find(ll x){
        if(x != p[x]) p[x] = find(p[x]);
        return p[x];
    }

    bool same(ll x, ll y){
        return find(x) == find(y);
    }

    void unite(ll x, ll y){
        x = find(x);
        y = find(y);
        if(x == y) return;
        if(rank[x] > rank[y]){
            p[y] = x;
            cnt[x] += cnt[y];
        }else{
            p[x] = y;
            cnt[y] += cnt[x];
            if(rank[x] == rank[y]) rank[y]++;
        }
    }

    ll size(ll x){
        return cnt[find(x)];
    }
};

int main(){
    ll n; cin >> n;
    vpl a(n),b(n);
    rep(i,n){
        cin >> a[i].first;
        a[i].second = i;
    }
    rep(i,n){
        cin >> b[i].first;
        b[i].second = i;
    }
    sort(all(a)); sort(all(b));
    rep(i,n){
        if(a[i].first>b[i].first){
            puts("No");
            return 0;
        }
        a[i].first = i;
        b[i].first = i;
    }
    rep(i,n){
        swap(a[i].first,a[i].second);
        swap(b[i].first, b[i].second);
    }
    sort(all(a)); sort(all(b));
    UnionFind uf(n);
    rep(i,n){
        uf.unite(a[i].first, b[i].first);
    }
    rep(i,n){
        if(uf.size(i)==n){
            puts("No");
            return 0;
        }
    }
    puts("Yes");
}