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 rep(i,n) for (int i = 0; i< (n); i++)
using ll = long long;
using namespace std;

int main(void){
	int N;
	cin >> N;
	int A[N];
	vector<int> cnt(N+1);
	int maxcnt = 1;
	rep(i,N) {
		cin >> A[i];
		cnt[A[i]]++;
	}
	int B[N];
	rep(i,N) {
		cin >> B[N-1-i];
		cnt[B[N-1-i]]++;
	}
	sort(cnt.begin(),cnt.end());
	maxcnt = cnt[N];

	if (maxcnt > N) {
		cout << "No" << endl;
		return 0;
	}

	int j = 0;
	rep(i,N) {
		while (A[i] == B[i]) {
			swap(B[i], B[j]);
			j++;
		}
	}
	j = N-1;
	rep(i,N) {
		while (A[i] == B[i]) {
			swap(B[i], B[j]);
			j--;
		}
	}

	cout << "Yes" << endl;
	rep(i,N) cout << B[i] << " ";
	cout << endl;

	return 0;
}
