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;

int a[200005], b[200005];

int main() {
    int n;
    scanf("%d", &n);

    for (int i = 0; i < n; i++) scanf("%d", &a[i]);
    for (int i = 0; i < n; i++) scanf("%d", &b[i]);

    reverse(b, b + n);

    int l = 0;
    int r = n - 1;

    for (int i = 0; i < n; i++) {
        if (a[i] != b[i]) continue;

        if (l < i && a[l] != b[i] && b[l] != a[i]) {
            swap(b[i], b[l]);
            l++;
        }
        else if (r > i && a[r] != b[i] && b[r] != a[i]) {
            swap(b[i], b[r]);
            r--;
        }
        else return 0 * printf("No\n");
    }

    printf("Yes\n");
    for (int i = 0; i < n; i++) {
        if (i == 0) printf("%d", b[i]);
        else printf(" %d", b[i]);
    }
    printf("\n");
}
