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 FOR(i, x, y) for(int i = (x); i < (y); ++i)
#define REP(i, x, y) for(int i = (x); i <= (y); ++i)
#define PB push_back
#define MP make_pair
#define PH push
#define fst first
#define snd second
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
typedef long double ldb;
typedef pair<int, int> pii;

const int maxn = 2e5 + 5;

int n;
int a[maxn], b[maxn];
vector<int> s, t;

int main(){
	scanf("%d", &n);
	FOR(i, 0, n)
		scanf("%d", a + i);
	FOR(i, 0, n)
		scanf("%d", b + i);
	reverse(b, b + n);
	
	int x = -1;
	FOR(i, 0, n) if(a[i] == b[i])
		s.PB(i), x = a[i]; 
	FOR(i, 0, n) if(a[i] != x && b[i] != x)
		t.PB(i);
	if(s.size() > t.size()){
		puts("No");
		return 0;
	}
		
	FOR(i, 0, s.size())
		swap(b[s[i]], b[t[i]]);
	puts("Yes");
	FOR(i, 0, n)
		printf("%d ", b[i]);
	puts("");
	return 0;
}
