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;
#define ll long long int
#define rep(i, n) for (int i = 0; i < (n); ++i)
#define FOR(i, a, b) for (int i = a; i < (b); ++i)
#define vl vector<ll>

int main() {

    ll N;
    cin >> N;

    vl A(N);
    vl B(N);
    vl a(N+1, 0);
    vl b(N+1, 0);

    rep(i, N) {
        cin >> A[i];
        a[A[i]]++;
    }

    rep(i, N) {
        cin >> B[i];
        b[B[i]]++;
    }

    FOR(i, 1, N+1) {
        if (a[i] + b[i] > N) {
            cout << "No" << endl;
            return 0;
        }
    }

    cout << "Yes" << endl;
    FOR(i, 1, N+1) {
        a[i] += a[i-1];
        b[i] += b[i-1];
    }

    ll k = 0;
    FOR(i, 1, N+1) {
        k = max(k, a[i]-b[i-1]);
    }

    rep(i, N) {
        cout << B[(i+N-k)%N] << " ";
    }
    cout << endl;

    return 0;
}
