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;
const int INF = 0x3f3f3f3f;
const LL mod = 1e9 + 7;
const int N = 200005;
int a[N], b[N], c[N];
int vis[N];
int main() {
    int n;
    cin >> n;
    for (int i = 0; i < n; i++) cin >> a[i], c[a[i]]++;
    for (int i = 0; i < n; i++) cin >> b[i], c[b[i]]++;
    for (int i = 1; i <= n; i++) {
        if (c[i] > n) {
            return puts("No"), 0;
        }
    }
    reverse(b, b + n);
    int pos = 0;
    for (int i = 0; i < n; i++) {
        if (a[i] == b[i]) {
            vis[i] = 1;
            while (a[pos] == a[i] || b[pos] == a[i] || vis[pos]) pos++;
            swap(b[pos], b[i]);
            pos++;
        }
    }
    puts("Yes");
    for (int i = 0; i < n; i++) printf("%d%c", b[i], " \n"[i + 1 == n]);
    return 0;
}
