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 int long long 


const int N = 2e5 + 7;

int a[N], b[N];

int32_t main() {
	ios_base :: sync_with_stdio(0); cin.tie(0); cout.tie(0);
	int n; cin >> n;
	for(int i = 1; i <= n; i++) cin >> a[i];
	for(int i = 1; i <= n; i++) cin >> b[i];

	reverse(b + 1, b + n + 1);

	vector <int> x, y;

	int z = -1;
	for(int i = 1; i <= n; i++) {
		if(a[i] == b[i]) {
			z = b[i];
			x.push_back(i);
		}
	}

	for(int i = 1; i <= n && y.size() < x.size(); i++) if((a[i] ^ z) && (b[i] ^ z)) {
		y.push_back(i);
	}

	if(x.size() > y.size()) {
		cout << "No" << endl;
		return 0;
	}

	cout << "Yes" << endl;
	for(int i = 0; i < x.size(); i++) swap(b[x[i]], b[y[i]]);
	for(int i = 1; i <= n; i++) cout << b[i] << " \n"[i == n];
}