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 < n; i++)
#define rrep(i, n) for (int i = n - 1; i >= 0; i--)
using namespace std;
#define INF ((1<<30)-1)
#define LINF (1LL<<60)
#define EPS (1e-10)
typedef long long ll;
typedef pair<ll, ll> P;
const int MOD = 1000000007;
const int MOD2 = 998244353;

int a[200010], b[200010], c[200010], d[200010];
map<int, int> mpa, mpb;

int main(){
    int n;
    cin >> n;
    rep(i, n) cin >> a[i], mpa[a[i]]++, c[a[i]]++;
    rep(i, n) cin >> b[i], mpb[b[i]]++, d[b[i]]++;
    rep(i, n) c[i+1] += c[i], d[i+1] += d[i];

    int idx = 0;
    for(auto e : mpa){
        if (e.second + mpb[e.first] > n){
            cout << "No" << endl;
            return 0;
        }
    }

    rep(i, n) idx = max(idx, c[i+1] - d[i]);

    cout << "Yes" << endl;

    rep(i, n){
        cout << b[(i+n-idx)%n] << " ";
    }
    cout << endl;
    return 0;
}
