Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
int numa[1000000], numb[1000000];
int main(){
    int n; cin >> n;
    vector<int> a, b;
    int x, y;
    for (int i = 0; i < n; i++)
    {
        cin >> x;
        a.push_back(x);
        numa[x]++;
    }
    for (int i = 0; i < n; i++)
    {
        cin >> y;
        b.push_back(y);
        numb[y]++;
    }
    for (int i = 1; i <= n; i++)
    {
        if(numa[i]+numb[i] > n) {
            puts("No");
            return 0;
        }
    }
    puts("Yes");
    for (int i = 1; i <= n; i++)
    {
        numa[i] += numa[i-1];
        numb[i] += numb[i-1];
    }
    int shift=0;
    for (int i = 1; i <= n; i++)
    {
        shift = max(shift, numa[i] - numb[i-1]);
    }
    for(int i = 0; i < n; i++){
        cout << b.at((i+n-shift)%n) << " ";
    }
    puts("");
}