#include <iostream>
#include <vector>
#include <utility>
#include <map>
#include <algorithm>
#include <queue>
#include <cmath>

using namespace std;

typedef pair<int,int> P;

int main() {
    int n;
    bool ans = false;
    cin >> n;
    vector<P> p(n);
    vector<int> a(n);
    for (int i=0; i<n; i++) {
        cin >> a[i];
        p[i].second = a[i];
    }

    for (int i=0; i<n; i++) cin>>p[i].first;
    sort(p.begin(), p.end());
    sort(a.begin(), a.end());

    for (int i=0; i<n; i++) {
        if (a[i] > p[i].first) {
            puts("No");
            return 0;
        }
    }


    for (int i=0; i<n-1; i++)
        if (a[i+1] <= p[i].first) ans = true;

    map<int,int> trans;
    for (int i=0; i<n; i++) trans[p[i].second] = i;

    int i=0;
    vector<bool> used(n,false);
    int cnt = 0;
    while (!used[i]) {
        used[i] = true;
        i = trans[a[i]];
        cnt++;
    }
    if (cnt != n) ans = true;

    cout << (ans? "Yes" : "No") << endl;
}
### Prompt

Please provide a cpp coded solution to the problem described below:

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`.

Examples

Input

3
1 3 2
1 2 3


Output

Yes


Input

3
1 2 3
2 2 2


Output

No


Input

6
3 1 2 6 3 4
2 2 8 3 4 3


Output

Yes

### Response

```cpp
#include <iostream>
#include <vector>
#include <utility>
#include <map>
#include <algorithm>
#include <queue>
#include <cmath>

using namespace std;

typedef pair<int,int> P;

int main() {
    int n;
    bool ans = false;
    cin >> n;
    vector<P> p(n);
    vector<int> a(n);
    for (int i=0; i<n; i++) {
        cin >> a[i];
        p[i].second = a[i];
    }

    for (int i=0; i<n; i++) cin>>p[i].first;
    sort(p.begin(), p.end());
    sort(a.begin(), a.end());

    for (int i=0; i<n; i++) {
        if (a[i] > p[i].first) {
            puts("No");
            return 0;
        }
    }


    for (int i=0; i<n-1; i++)
        if (a[i+1] <= p[i].first) ans = true;

    map<int,int> trans;
    for (int i=0; i<n; i++) trans[p[i].second] = i;

    int i=0;
    vector<bool> used(n,false);
    int cnt = 0;
    while (!used[i]) {
        used[i] = true;
        i = trans[a[i]];
        cnt++;
    }
    if (cnt != n) ans = true;

    cout << (ans? "Yes" : "No") << endl;
}
```