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;
long long mod = 1000000007;
int main() {
    long long n, c[200005], d[200005], x;
    cin >> n;
    long long a[n], b[n];
    for (long long i = 0; i < 200005; i++) 
    {
        c[i] = 0;
        d[i] = 0;
    }
    for (long long i = 0; i < n; i++)
    {
        cin >> a[i];
        c[a[i]]++;
    }
    for (long long i = 0; i < n; i++)
    {
        cin >> b[i];
        d[b[i]]++;
    }
    for (long long i = 0; i < 200005; i++)
    {
        if (c[i] + d[i] > n)
        {
            cout << "No";
            return 0;
        }
    }
    cout << "Yes" << endl;
    for (int i = 1; i <= n; i++)
    {
	    c[i] += c[i - 1];
	    d[i] += d[i - 1];
    }
    for (int i = 1; i <= n; i++) x = max(x, c[i] - d[i - 1]);
    for (int i = 0; i < n; i++)
    	cout << b[(i + n - x) % n] << ' ';
}
