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 acnt[200010];
int bcnt[200010];
int main() {
	int n; cin >> n;
	int a[n]; for(int i = 0; i < n; ++i) cin >> a[i];
	int b[n]; for(int i = 0; i < n; ++i) cin >> b[i];
	for(int i = 0; i < n; ++i) {
		++acnt[a[i]];
		++bcnt[b[i]];
	}
	for(int i = 1; i <= n; ++i) {
		if(acnt[i] + bcnt[i] > n) {
			cout << "No" << '\n';
			return 0;
		}
		acnt[i] += acnt[i - 1];
		bcnt[i] += bcnt[i - 1];
	}
	int maxi = 0;
	for(int i = 1; i <= n; ++i) {
		maxi = max(maxi, acnt[i] - bcnt[i - 1]);
	}
	int ans[n];
	for(int i = 0; i < n; ++i) {
		ans[(i + maxi) % n] = b[i];
	}
	cout << "Yes" << '\n';
	for(int i = 0; i < n; ++i) {
		cout << ans[i] << (i == n - 1 ? "\n" : " ");
	}
	return 0;
}
