#include <iostream>
#include <vector>
#include <utility>
#include <algorithm>
using namespace std;

int main()
{
    int N;
    cin>>N;
    vector<int> A(N), B(N);
    for (int &a: A)
        cin>>a;
    for (int &b: B)
        cin>>b;

    vector<pair<int,int>> T;
    for (int i=0; i<N; i++)
        T.push_back(make_pair(B[i], A[i]));
    sort(T.begin(), T.end());
    for (int i=0; i<N; i++)
        A[i] = T[i].second,
        B[i] = T[i].first;

    vector<int> I(N);
    for (int i=0; i<N; i++)
        I[i] = i;
    sort(I.begin(), I.end(), [&](int a, int b){return A[a]<A[b];});

    int c = 1;
    int p = I[0];
    while (p!=0)
    {
        p = I[p];
        c++;
    }

    bool ok = true;
    for (int i=0; i<N; i++)
        if (A[I[i]]>B[i])
            ok = false;
    if (!ok)
    {
        cout<<"No"<<endl;
        return 0;
    }

    if (c<N)
    {
        cout<<"Yes"<<endl;
        return 0;
    }

    for (int i=0; i<N-1; i++)
        if (A[I[i+1]]<=B[i] && A[I[i]]<=B[i+1])
        {
            cout<<"Yes"<<endl;
            return 0;
        }

    cout<<"No"<<endl;
}

### Prompt

Your challenge is to write a Cpp solution to the following problem:

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 <algorithm>
using namespace std;

int main()
{
    int N;
    cin>>N;
    vector<int> A(N), B(N);
    for (int &a: A)
        cin>>a;
    for (int &b: B)
        cin>>b;

    vector<pair<int,int>> T;
    for (int i=0; i<N; i++)
        T.push_back(make_pair(B[i], A[i]));
    sort(T.begin(), T.end());
    for (int i=0; i<N; i++)
        A[i] = T[i].second,
        B[i] = T[i].first;

    vector<int> I(N);
    for (int i=0; i<N; i++)
        I[i] = i;
    sort(I.begin(), I.end(), [&](int a, int b){return A[a]<A[b];});

    int c = 1;
    int p = I[0];
    while (p!=0)
    {
        p = I[p];
        c++;
    }

    bool ok = true;
    for (int i=0; i<N; i++)
        if (A[I[i]]>B[i])
            ok = false;
    if (!ok)
    {
        cout<<"No"<<endl;
        return 0;
    }

    if (c<N)
    {
        cout<<"Yes"<<endl;
        return 0;
    }

    for (int i=0; i<N-1; i++)
        if (A[I[i+1]]<=B[i] && A[I[i]]<=B[i+1])
        {
            cout<<"Yes"<<endl;
            return 0;
        }

    cout<<"No"<<endl;
}

```