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

typedef long long ll;

int main() {
    int N;
    cin >> N;
    vector<int> A(N), B(N), C(N + 1, 0), D(N + 1, 0), E(N + 1, 0), F(N + 1, 0);
    rep(i, N) {
        cin >> A[i];
        E[A[i]]++;
    }
    rep(i, N) {
        cin >> B[i];
        F[B[i]]++;
    }
    rep(i, N) {
        if (E[i + 1] + F[i + 1] > N) {
            cout << "No\n";
            return 0;
        }
    }
    cout << "Yes\n";
    rep(i, N) {
        C[i + 1] = C[i] + E[i + 1];
        D[i + 1] = D[i] + F[i + 1];
    }
    int x = -N;
    rep(i, N) {
        x = max(x, C[i + 1] - D[i]);
    }
    rep(i, N) {
        cout << B[(i - x + N) % N];
        if (i == N - 1) cout << "\n";
        else cout << " ";
    }
}