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 rep(i, n) for (int i = 0; i < (int)(n); i++)
using ll = long long;
using P = pair<int, int>;

#define INF 1001001001
#define MAX 200005

int main() {
  int n;
  cin >> n;
  vector<int> a(n), b(n), c(n+1), d(n+1);
  rep(i,n) cin >> a[i], c[a[i]]++;
  rep(i,n) cin >> b[i], d[b[i]]++;
  rep(i,n+1) if (c[i]+d[i]>n) {
    cout << "No" << endl;
    return 0;
  }

  rep(i,n) c[i+1]+=c[i],d[i+1]+=d[i];
  int x = 0;
  rep(i,n) x = max(x,c[i+1]-d[i]);

  cout << "Yes" << endl;
  rep(i,n) {
    cout << b[(i+n-x)%n];
    if (i==n-1) cout << endl; else cout << " ";
  }

  return 0;
}
