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;

typedef long long ll;
typedef pair<int,int> P;

#define p_ary(ary,a,b) do { cout << "["; for (int count = (a);count < (b);++count) cout << ary[count] << ((b)-1 == count ? "" : ", "); cout << "]\n"; } while(0)
#define p_map(map,it) do {cout << "{";for (auto (it) = map.begin();;++(it)) {if ((it) == map.end()) {cout << "}\n";break;}else cout << "" << (it)->first << "=>" << (it)->second << ", ";}}while(0)

int main() {
    int n;
    cin >> n;
    vector<P> a(n),b(n);
    vector<int> p(n);
    for (int i = 0;i < n;++i) {
        cin >> a[i].first;
        a[i].second = i;
    }
    for (int i = 0;i < n;++i) {
        cin >> b[i].first;
        b[i].second = i;
    }
    sort(a.begin(),a.end());
    sort(b.begin(),b.end());
    for (int i = 0;i < n;++i) if (a[i].first > b[i].first) {
        cout << "No" << endl;
        return 0;
    }
    vector<bool> used(n,false);
    for (int i = 0;i < n;++i) p[a[i].second] = b[i].second;
    for (int i = 0;i < n;++i) if (!used[i]) {
        if (i) {
            cout << "Yes" << endl;
            return 0;
        }
        int j = 0;
        while (!used[j]) {
            used[j] = true;
            j = p[j];
        }
    }
    for (int i = 0;i < n-1;++i) if (a[i+1].first <= b[i].first) {
        cout << "Yes" << endl;
        return 0;
    }
    cout << "No" << endl;
    return 0;
}