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;
const int maxn = 400001;
int n, a[maxn], b[maxn], ta[maxn], tb[maxn], l = 1, r = 1;
int main() {
	cin >> n;
	for(int i = 1; i <= n; ++ i) {
		cin >> a[i];
		ta[a[i]] ++;
	}
	for(int i = 1; i <= n; ++ i) {
		cin >> b[i];
		tb[b[i]]++;
	}
	for(int i = 1; i <= n; ++ i) {
		if(ta[i] + tb[i] > n) {
			puts("No");
			exit(0);
		}
	}
	memset(b, 0, sizeof(b));
	for(int i = 1;i <= n; ++ i) {
		for(int j = 1; j <= tb[i]; ++ j) {
			while(b[l]) ++ l;
			while(a[r] <= a[l] && r <= n) ++ r;
			while(b[r]) ++ r;
			if(a[l] != i) b[l ++] = i;
			else b[r ++] = i;
		}
	}
	l = 1;
	for(int i = n + 1; b[i]; ++ i) {
		while(b[l]) ++ l;
		if(a[l] != b[i]) a[l] = b[i];
		else {
			while(a[r = (rand() % n + 1)] == b[i] || b[r] == a[l]);
			b[l] = b[r];
			b[r] = b[i];
		}
	}
	puts("Yes");
	for(int i = 1; i <= n; ++ i) printf("%d ", b[i]);
	return 0;
}
