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;
template <class T>
inline bool chmax(T &a, T b)
{
    if (a < b)
    {
        a = b;
        return 1;
    }
    return 0;
}
template <class T>
inline bool chmin(T &a, T b)
{
    if (a > b)
    {
        a = b;
        return 1;
    }
    return 0;
}
typedef long long int ll;

#define ALL(v) (v).begin(), (v).end()
#define RALL(v) (v).rbegin(), (v).rend()
#define endl "\n"
const double EPS = 1e-7;
const int INF = 1 << 30;
const ll LLINF = 1LL << 60;
const double PI = acos(-1);
const int MOD = 1000000007;
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};

//-------------------------------------
using P = pair<ll, ll>;

struct UnionFind
{
    vector<int> par;

    UnionFind(int n) : par(n, -1) {}
    void init(int n) { par.assign(n, -1); }

    int root(int x)
    {
        if (par[x] < 0)
            return x;
        else
            return par[x] = root(par[x]);
    }

    bool issame(int x, int y)
    {
        return root(x) == root(y);
    }

    bool merge(int x, int y)
    {
        x = root(x);
        y = root(y);
        if (x == y)
            return false;
        if (par[x] > par[y])
            swap(x, y); // merge technique
        par[x] += par[y];
        par[y] = x;
        return true;
    }

    int size(int x)
    {
        return -par[root(x)];
    }
};

int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    ll n;
    cin >> n;
    vector<P> a(n), b(n);
    for (ll i = 0; i < n; i++)
    {
        cin >> a[i].first;
        a[i].second = i;
    }
    for (ll i = 0; i < n; i++)
    {
        cin >> b[i].first;
        b[i].second = i;
    }
    sort(ALL(a));
    sort(ALL(b));
    UnionFind uf(n);
    bool flg = false;
    for (ll i = 0; i < n; i++)
    {
        if (a[i].first > b[i].first)
        {
            cout << "No" << endl;
            return 0;
        }
        uf.merge(a[i].second, b[i].second);
        if (i > 0 && a[i].first <= b[i - 1].first)
        {
            flg = true;
        }
    }
    if (uf.size(0) != n || flg)
    {
        cout << "Yes" << endl;
    }
    else
    {
        cout << "No" << endl;
    }
}