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<iostream>
#include<algorithm>

using namespace std;
typedef long long ll;
const int N = 200010;
int n, a[N], b[N], cnt, tot;

int main() {
    cin >> n;
    for (int i = 1; i <= n; i++) cin >> a[i];
    for (int i = 1; i <= n; i++) cin >> b[n - i + 1];
    int l = 1, r = n, st = 0, ed = -1;
    for (int i = 1; i <= n; i++)
        if (a[i] == b[i]) {
            if (!st) st = i;
            ed = i;
        }
    for (int j = st; j <= ed; j++) {
        if (a[l] ^ b[j] && b[l] ^ b[j]) swap(b[j], b[l++]);
        else if (a[r] ^ b[j] && b[r] ^ b[j]) swap(b[j], b[r--]);
        else puts("No"), exit(0);
    }
    puts("Yes");
    for (int i = 1; i <= n; i++) printf("%d%c", b[i], " \n"[i == n]);
    return 0;
}
