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;
typedef long long ll;

int main() {
	int n;
    cin >> n;
    vector<int> a(n), b(n), counter(n+1, 0);
    vector<int> c(n+1, 0), d(n+1, 0);
    for (int i = 0; i < n; ++i) {
        cin >> a[i];
        counter[a[i]]++;
        c[a[i]]++;
    }
    for (int i = 0; i < n; ++i) {
        cin >> b[i];
        counter[b[i]]++;
        d[b[i]]++;
    }

    for (int i = 0; i <= n; ++i) {
        if(counter[i] > n) {
            cout << "No" << endl;
            return 0;
        }
    }
    cout << "Yes" << endl;

    for (int i = 0; i < n; ++i) {
        c[i+1] += c[i];
        d[i+1] += d[i];
    }

    int ans = 0;
    for (int i = 1; i <= n; ++i) {
        ans = max(ans, c[i] - d[i-1]);
    }

    for (int i = 0; i < n; ++i) {
        cout << b[(i + n - ans) % n] << " ";
    }
    cout << endl;
}
