There are n children numbered from 1 to n in a kindergarten. Kindergarten teacher gave a_i (1 ≤ a_i ≤ n) candies to the i-th child. Children were seated in a row in order from 1 to n from left to right and started eating candies. 

While the i-th child was eating candies, he calculated two numbers l_i and r_i — the number of children seating to the left of him that got more candies than he and the number of children seating to the right of him that got more candies than he, respectively.

Formally, l_i is the number of indices j (1 ≤ j < i), such that a_i < a_j and r_i is the number of indices j (i < j ≤ n), such that a_i < a_j.

Each child told to the kindergarten teacher the numbers l_i and r_i that he calculated. Unfortunately, she forgot how many candies she has given to each child. So, she asks you for help: given the arrays l and r determine whether she could have given the candies to the children such that all children correctly calculated their values l_i and r_i, or some of them have definitely made a mistake. If it was possible, find any way how she could have done it.

Input

On the first line there is a single integer n (1 ≤ n ≤ 1000) — the number of children in the kindergarten.

On the next line there are n integers l_1, l_2, …, l_n (0 ≤ l_i ≤ n), separated by spaces.

On the next line, there are n integer numbers r_1, r_2, …, r_n (0 ≤ r_i ≤ n), separated by spaces.

Output

If there is no way to distribute the candies to the children so that all of them calculated their numbers correctly, print «NO» (without quotes).

Otherwise, print «YES» (without quotes) on the first line. On the next line, print n integers a_1, a_2, …, a_n, separated by spaces — the numbers of candies the children 1, 2, …, n received, respectively. Note that some of these numbers can be equal, but all numbers should satisfy the condition 1 ≤ a_i ≤ n. The number of children seating to the left of the i-th child that got more candies than he should be equal to l_i and the number of children seating to the right of the i-th child that got more candies than he should be equal to r_i. If there is more than one solution, find any of them.

Examples

Input

5
0 0 1 1 2
2 0 1 0 0


Output

YES
1 3 1 2 1


Input

4
0 0 2 0
1 1 1 1


Output

NO


Input

3
0 0 0
0 0 0


Output

YES
1 1 1

Note

In the first example, if the teacher distributed 1, 3, 1, 2, 1 candies to 1-st, 2-nd, 3-rd, 4-th, 5-th child, respectively, then all the values calculated by the children are correct. For example, the 5-th child was given 1 candy, to the left of him 2 children were given 1 candy, 1 child was given 2 candies and 1 child — 3 candies, so there are 2 children to the left of him that were given more candies than him.

In the second example it is impossible to distribute the candies, because the 4-th child made a mistake in calculating the value of r_4, because there are no children to the right of him, so r_4 should be equal to 0.

In the last example all children may have got the same number of candies, that's why all the numbers are 0. Note that each child should receive at least one candy.
import java.util.*;
public class MyClass {
    public static void main(String args[]) {
      Scanner sc=new Scanner(System.in);
      int n=sc.nextInt();
      int flag=0;
      int le[]=new int[n];
      int ri[]=new int[n];
      int ans[]=new int[n];
      for(int i=0;i<n;i++)
      {
          le[i]=sc.nextInt();
      }
      for(int i=0;i<n;i++)
      {
          ri[i]=sc.nextInt();
      }
      boolean vis[]=new boolean[n];
      if(le[0]!=0 || ri[n-1]!=0)
      {
          flag=1;
      }
      int ind=0;
      int value=n;
      int cnt=0,once=0;
      //System.out.println(flag);
      ArrayList<Integer> al=new ArrayList<>();
      if(flag!=1)
      {
        while(cnt<n && flag!=1)
        {
            once=0;
            al.clear();
            for(int i=0;i<n;i++)
            {
                if(le[i]==0 && ri[i]==0 && vis[i]==false)
                {
                    vis[i]=true;
                    ans[i]=value;
                    ind=i;
                    al.add(ind);
                    //System.out.println(ind);
                    cnt++;
                    once=1;
                    
                }
            }
            for(int i=0;i<al.size();i++)
            {
                    ind=al.get(i);
                    for(int j=ind-1;j>=0;j--)
                    {
                        if(vis[j]==false)
                        {
                            ri[j]--;
                        }
                    }
                    for(int j=ind+1;j<n;j++)
                    {
                        if(vis[j]==false)
                            le[j]--;
                    }
            }
            if(once==0)
            {
                flag=1;
                break;
            }
            value--;
        }
      }
      if(flag==1)
      {
          System.out.println("NO");
      }
      else
      {
          System.out.println("YES");
          for(int i=0;i<n;i++)
          {
              System.out.print(ans[i]+" ");
          }
      }
    }
}