Problem: 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

Solution (in CPP):

Solution:
#include "bits/stdc++.h"
using namespace std;
#include "string"
#define int long long
#define pi pair <int, int>
#define ff first
#define ss second
#define boost ios::sync_with_stdio(false);cin.tie(nullptr)
#define endl '\n'

int32_t main() {
	boost;
	int n;
	cin >> n;
	vector < int > a(n), b(n);
	for(int &i : a)
		cin >> i;
	for(int &i : b)
		cin >> i;
	reverse(b.begin(), b.end());
	int l = 0, r = 0;
	for(int i = 0; i < n; i++) {
		if(a[i] != b[i])
			continue;
		l = i, r = i;
		while(r < n && a[r] == b[r])
			r++;
		break;
	}
	r--;

	for(int i = 0; i < n && l <= r; i++) {
		if(b[i] != a[l] && a[i] != b[l])
			swap(b[i], b[l++]);
	}
	if(l <= r) {
		cout << "No" << endl;
		return 0;
	}

	cout << "Yes" << endl;
	for(int &i : b)
		cout << i << ' ';
	cout << endl;
}