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 n, a[200005], b[200005];

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

//    freopen("input.txt","r",stdin);
//    freopen("output.txt","w",stdout);

    cin >> n;
    for (int i = 0; i < n; i++) {
        cin >> a[i];
    }
    for (int i = 0; i < n; i++) {
        cin >> b[i];
    }

    reverse(b, b + n);

    int l = 0, r = n - 1;
    while (l < n && a[l] != b[l]) l++;
    while (r >= 0 && a[r] != b[r]) r--;

    int i, j;
    for (i = l, j = 0; i <= r && j < n; j++) {
        if (a[j] != b[i] && b[j] != b[i]) {
            swap(b[j], b[i]);
            i++;
        }
    }

    if (l > r || i == r + 1) {
        cout << "Yes\n";
        for (int i = 0; i < n; i++) {
            cout << b[i] << " \n"[i == n - 1];
        }
    }
    else cout << "No\n";

    return 0;
}
