Given are two sequences A and B, both of length N. A and B are each sorted in the ascending order. Check if it is possible to reorder the terms of B so that for each i (1 \leq i \leq N) A_i \neq B_i holds, and if it is possible, output any of the reorderings that achieve it.

Constraints

* 1\leq N \leq 2 \times 10^5
* 1\leq A_i,B_i \leq N
* A and B are each sorted in the ascending order.
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N
A_1 A_2 \cdots A_N
B_1 B_2 \cdots B_N


Output

If there exist no reorderings that satisfy the condition, print `No`.

If there exists a reordering that satisfies the condition, print `Yes` on the first line. After that, print a reordering of B on the second line, separating terms with a whitespace.

If there are multiple reorderings that satisfy the condition, you can print any of them.

Examples

Input

6
1 1 1 2 2 3
1 1 1 2 2 3


Output

Yes
2 2 3 1 1 1


Input

3
1 1 2
1 1 3


Output

No


Input

4
1 1 2 3
1 2 3 3


Output

Yes
3 3 1 2
#include <bits/stdc++.h>

using namespace std;
using vi = vector<int>;

int main()
{
    int n; cin >> n;
    vi a(n), b(n), ind_a(n + 1, -1), sum(n + 1);
    int k = 0;
    for(int i = 0; i < n; ++i){
        cin >> a[i];
        ind_a[a[i]] = i;
        sum[a[i]]++;
    }
    for(int i = 0; i < n; ++i){
        cin >> b[i];
        if((!i || b[i] != b[i - 1]) && ind_a[b[i]] != -1){
            k = max(k, ind_a[b[i]] - i + 1);
        }
        sum[b[i]]++;
    }
    for(int i = 0; i <= n; ++i){
        if(sum[i] > n){
            cout << "No";
            return 0;
        }
    }
    cout << "Yes\n";
    for(int i = n - k; i < n; ++i){
        cout << b[i] << " ";
    }
    for(int i = 0; i < n - k; ++i)
        cout << b[i] << " ";
    return 0;
}
