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;
using ll = long long;

int main() {
	ll N;
	cin >> N;
	vector<ll> a(N), b(N);
	for(ll i=0; i<N; i++) cin >> a[i];
	for(ll i=0; i<N; i++) cin >> b[i];

	map<ll,ll> am,bm;
	for(ll i=0; i<N; i++){
		am[a[i]]++;
		bm[b[i]]++;
	}
	for(auto&& pr:am){
		if(pr.second + bm[pr.first]> N){
			cout << "No\n";
			return 0;
		}
	}
	
	reverse(b.begin(), b.end());
	ll index=0;
	for(ll i=0; i<N; i++){
		if(a[i]==b[i]){
			while(a[index]==a[i] || b[index]==a[i])index++;
			swap(b[index],b[i]);
			index++;
		}
	}

	cout << "Yes\n";
	for(ll i=0; i<N; i++) cout << b[i] << " \n"[i==N-1];

	return 0;
}
