Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e5 + 10;

int a[MAXN], b[MAXN];
int main() {
    int n;
    cin >> n;
    map<int, int> m;
    for (int i = 0; i < n; i++) {
        cin >> a[i];
        m[a[i]]++;
    }
    for (int i = 0; i < n; i++) {
        cin >> b[i];
        m[b[i]]++;
    }
    for (auto it : m) {
        if (it.second > n) {
            cout << "No\n";
            return 0;
        }
    }
    reverse(b, b + n);
    int l = 0, r = n - 1;
    for (int i = 0; i < n; i++) {
        if (a[i] == b[i]) {
            if(a[i] != b[l] and a[l] != b[i]) {
                swap(b[l], b[i]);
                l ++;
            }
            else {
                swap(b[r], b[i]);
                r --;
            }
        }
    }
    cout << "Yes\n";
    for (int i = 0; i < n; i++) cout << b[i] << " ";
    cout << endl;
}