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;
using LL = long long;
using ULL = unsigned long long;
#define rep(i,n) for(int i=0; i<(n); i++)

int N;
int A[200001];
int B[200001];
int cA[200001]={};
int cB[200001]={};
int ans[200001];

int main(){
 scanf("%d",&N);
 rep(i,N){ scanf("%d",&A[i]); cA[A[i]]++; A[i]--; }
 rep(i,N){ scanf("%d",&B[i]); cB[B[i]]++; B[i]--; }
 A[N] = B[N] = 1000000000;

 rep(i,N){ if(cA[i+1]+cB[i+1]>N){ printf("No\n"); return 0; } }
 rep(i,N){ cA[i+1]+=cA[i]; cB[i+1]+=cB[i]; }

 int d = 0;
 rep(i,N) d = max(d,cA[i+1]-cB[i]);

 printf("Yes\n");
 rep(i,N){
  if(i) printf(" ");
  printf("%d",B[(i+N-d)%N]+1);
 } printf("\n");

 return 0;
}


