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>

#define debug(...) fprintf(stderr, __VA_ARGS__)

#ifndef AT_HOME
#define getchar() IO::myGetchar()
#define putchar(x) IO::myPutchar(x)
#endif
using namespace std;

int n;
int a[200005], b[200005];

int main() {
	cin >> n;
	for (int i = 1; i <= n; ++i) {
    cin >> a[i];
	}
	for (int i = 1; i <= n; ++i) {
    cin >> b[i];
	}
	int d = 0;
	int pa = 1, pb = 1;
	for (int i = 1; i <= n; ++i) {
		while (pa <= n && a[pa] == i) {
			++pa;
		}
		d = std::max(d, pa - pb);
		while (pb <= n && b[pb] == i) {
			++pb;
		}
	}
	std::rotate(b + 1, b + n - d + 1, b + 1 + n);
	for (int i = 1; i <= n; ++i) {
		if (a[i] == b[i]) {
      cout << "No\n";
			return 0;
		}
	}
  cout << "Yes\n";
	for (int i = 1; i <= n; ++i) {
    cout << b[i] << " ";
	}
  cout << "\n";
}

