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 rep(i,n) for(int i=0;i<n;i++)
#define REP(i,n) for(int i=1;i<=n;i++)
typedef long long ll;

int main(){
	int n;
	cin >> n;
	vector<int> a(n);
	vector<int> b(n);
	vector<int> acount(n+1);
	vector<int> bcount(n+1);
	rep(i,n){
		cin >> a[i];
		acount[a[i]]++;
	}
	rep(i,n){
		cin >> b[i];
		bcount[b[i]]++;
	}
	REP(i,n){
		if(acount[i] + bcount[i] > n){
			cout << "No" << endl;
			return 0;
		}
	}
	sort(b.rbegin(),b.rend());
	queue<int> que;
	rep(i,n){
		if(a[i] == b[i]) que.push(i);
	}
	srand((unsigned int)time(NULL));
	while(!que.empty()){
		int p = que.front(); que.pop();
		int idx = rand() % n;
		swap(b[p], b[idx]);
		if(b[p] == a[p]) que.push(p);
		if(b[idx] == a[idx]) que.push(idx);
	}
	cout << "Yes" << endl;
	rep(i,n) cout << b[i] << " ";
	cout << endl;
    return 0;
}