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 <iostream>
#include <string>
#include <vector>
#include <set>
#include <queue>
#include <stack>
#include <map>
#include <algorithm>
#include <math.h>
#include <cassert>
#define rep(i,n) for(int i = 0; i < n; ++i )
using namespace std;
using ll = long long;
using P = pair<int,int>;

int main() {
  int n;
  cin >> n;
  vector<int> a(n),b(n);
  rep(i,n) cin >> a[i];
  rep(i,n) cin >> b[i];

  vector<int> s(n+1),t(n+1);
  rep(i,n) s[a[i]]++, t[b[i]]++;
  rep(i,n+1){
    if(s[i]+t[i]>n){
      cout << "No" << endl;
      return 0;
    }
  }
  rep(i,n) s[i+1]+=s[i], t[i+1]+=t[i];

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

  cout << "Yes\n";
  rep(i,n) cout << b[(i+n-x)%n] << (i==n-1?"\n":" ");
}
