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;
typedef long long ll;

int main () {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    int n; cin >> n;
    vector<int> a(n), b(n);
    for (int i = 0; i < n; i++) {
        cin >> a[i];
    }
    for (int i = 0; i < n; i++) {
        cin >> b[i];
    }
    reverse(b.begin(), b.end());
    int l = 0, r = n - 1;
    for (int i = 0; i < n; i++) {
        if (a[i] == b[i]) {
            if (b[i] != a[l] && b[i] != b[l]) {
                swap(b[i], b[l++]);
            }
            else if (b[i] != a[r] && b[i] != b[r]) {
                swap(b[i], b[r--]);
            }
            else {
                cout << "No" << endl;
                return 0;
            }
        }
    }
    cout << "Yes" << endl;
    for (int i : b) {
        cout << i << " ";
    }
    return 0;
}

