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>
#define rep(i, n) for (int i = 0; i < (n); i++)
using namespace std;

int main() {
    int n; cin >> n;

    vector<int> a(n), b(n);
    rep(i,n) cin >> a[i];
    rep(i,n) cin >> b[n- 1 - i];

    int save = 0;
    vector<int> ch1, ch2;
    rep(i,n) if (a[i] == b[i]) { save = a[i]; ch1.push_back(i); }
    rep(i,n) if (a[i] != save && b[i] != save) ch2.push_back(i);
    
    if ((int)ch1.size() > (int)ch2.size()) {
        cout << "No" << endl;
    }
    else {
        cout << "Yes" << endl;
        rep(i,(int)ch1.size()) swap(b[ch1[i]], b[ch2[i]]);
        rep(i,n) cout << b[i] << " ";
        cout << endl;
    }
    return 0;
}