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;

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);
	
	map<int,int> mp;
	int n ;
	cin >> n;
	vector<int> A(n), B(n);
	for(int i=0; i<n; i++) cin >> A[i];
	for(int i=0; i<n; i++) cin >> B[i];
	reverse(B.begin(), B.end());
	int el = -1;
	for(int i=0; i<n; i++){
		if(A[i] == B[i]){
			el = A[i];
			break;
		}
	}
	if(el == -1){
		cout << "Yes\n";
		for(int i=0; i<n; i++) cout << B[i] << " ";
		return 0;
	}
	int l, r;
	for(int i=0; i<n; i++){
		if(A[i]==el && B[i]==el){
			l = i;
			break;
		}
	}
	for(int i=n-1; i>=0; i--){
		if(A[i]==el && B[i]==el){
			r = i;
			break;
		}
	}
	for(int i=0; i<n; i++){
		if(l>r) break;
		if(A[i]!=el && B[i]!=el){
			swap(B[i], B[l]);
			l++;
		}
	}
	if(l<=r){
		cout << "No\n";
		return 0;
	}
	cout << "Yes\n";
	for(int i=0; i<n; i++) cout << B[i] << " ";
	return 0;
}