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;
using ll = long long;
int main()
{
    int n;
    cin >> n;
    vector<int> a(n);
    vector<int> b(n);
    vector<int> cnta(n+1, 0);
    vector<int> cntb(n+1, 0);
    for(int i = 0; i < n; i++)
    {
        cin >> a[i];
        cnta[a[i]]++;
    }
    for(int i = 0; i < n; i++)
    {
        cin >> b[i];
        cntb[b[i]]++;
    }
    for(int i = 0; i <= n; i++)
    {
        if(cnta[i] + cntb[i] > n)
        {
            cout << "No" << endl;
            return 0;
        }
    }
    for(int i = 1; i <= n; i++)
    {
        cnta[i] += cnta[i-1];
        cntb[i] += cntb[i-1];
    }
    cout << "Yes" << endl;
    int x = 0;
    for(int i = 1; i <= n; i++)
    {
        x = max(x, cnta[i] - cntb[i-1]);
    }
    for(int i = 0; i < n; i++)
    {
        cout << b[(i-x+n)%n] << " ";
    }
    cout << endl;
}
