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
/*input
6
1 1 1 2 2 3
1 1 1 2 2 3
*/

#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<int, int>;


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];

    vector<int> cnt(n+1);
    int mcnt = 0;
    rep(i, n) {
        mcnt = max({mcnt, ++cnt[a[i]], ++cnt[b[i]]});
    }
    if (mcnt > n) {
        cout << "No" << endl;
        return 0;
    }
    else cout << "Yes" << endl;
    int l = 0, r = n-1;
    rep(i, n) {
        if (a[i] == b[i]) {
            if (a[i] != b[l] && a[l] != b[i]) {
                swap(b[i], b[l++]);
            }
            else swap(b[i], b[r--]);
        }
    }
    // rep(i, n) cout << a[i] << " ";
    // cout << endl;
    rep(i, n) cout << b[i] << " ";
    cout << endl;
}