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.*;
import java.io.*;


public class Main {

    public static void main(String[] args) {
        InputReader in = new InputReader(System.in);
        OutputWriter out = new OutputWriter(System.out);
      int n = in.nextInt();
         int[]l = new int[n];
         int[]r = new int[n];
         for (int i =0;i<n;i++)
             l[i]=in.nextInt();
         for (int i=0;i<n;i++)
             r[i]=in.nextInt();
         int[]empty = new int[n];
         LazySegmentTree t = new LazySegmentTree();
         t.constructST(empty,n);
         boolean[]visited = new boolean[n];
        int[]ans = new int[n];

        for (int i =0;i<n;i++)
         {
             if (l[i]==0&&r[i]==0) {
                 t.updateRange(n, i, i, 1);
                 visited[i] = true;
                 ans[i]=n;
             }
         }
         int cntr=n-1;
         for (int j =0;j<n;j++)
         {
             List<Integer>list = new ArrayList<>();
             for (int i=0;i<n;i++)
             {

                 if (!visited[i]&&t.getSum(n,0,i-1)==l[i]&&t.getSum(n,i+1,n-1)==r[i])
                 {
                     list.add(i);
                     ans[i]=cntr;
                     visited[i]=true;
                 }
             }
             for (int i =0;i<list.size();i++)
                 t.updateRange(n,list.get(i),list.get(i),1);

             cntr--;
         }
         boolean check= true;
         for (int i =0;i<n;i++)
             if (!visited[i])
                 check=false;
         if (!check)
             out.printLine("NO");
         else {
             out.printLine("YES");
             for (int item : ans
                     ) {
                 out.print(item + " ");
             }
         }
        out.flush();
    }
}
class LazySegmentTree {
    final int MAX = 5000000;        // Max tree size
    int tree[] = new int[MAX];  // To store segment tree
    int lazy[] = new int[MAX];  // To store pending updates

    /*  si -> index of current node in segment tree
        ss and se -> Starting and ending indexes of elements for
                     which current nodes stores sum.
        us and eu -> starting and ending indexes of update query
        ue  -> ending index of update query
        diff -> which we need to add in the range us to ue */
    void updateRangeUtil(int si, int ss, int se, int us,
                         int ue, int diff) {
        // If lazy value is non-zero for current node of segment
        // tree, then there are some pending updates. So we need
        // to make sure that the pending updates are done before
        // making new updates. Because this value may be used by
        // parent after recursive calls (See last line of this
        // function)
        if (lazy[si] != 0) {
            // Make pending updates using value stored in lazy
            // nodes
            tree[si] += (se - ss + 1) * lazy[si];

            // checking if it is not leaf node because if
            // it is leaf node then we cannot go further
            if (ss != se) {
                // We can postpone updating children we don't
                // need their new values now.
                // Since we are not yet updating children of si,
                // we need to set lazy flags for the children
                lazy[si * 2 + 1] += lazy[si];
                lazy[si * 2 + 2] += lazy[si];
            }

            // Set the lazy value for current node as 0 as it
            // has been updated
            lazy[si] = 0;
        }

        // out of range
        if (ss > se || ss > ue || se < us)
            return;

        // Current segment is fully in range
        if (ss >= us && se <= ue) {
            // Add the difference to current node
            tree[si] += (se - ss + 1) * diff;

            // same logic for checking leaf node or not
            if (ss != se) {
                // This is where we store values in lazy nodes,
                // rather than updating the segment tree itelf
                // Since we don't need these updated values now
                // we postpone updates by storing values in lazy[]
                lazy[si * 2 + 1] += diff;
                lazy[si * 2 + 2] += diff;
            }
            return;
        }

        // If not completely in rang, but overlaps, recur for
        // children,
        int mid = (ss + se) / 2;
        updateRangeUtil(si * 2 + 1, ss, mid, us, ue, diff);
        updateRangeUtil(si * 2 + 2, mid + 1, se, us, ue, diff);

        // And use the result of children calls to update this
        // node
        tree[si] = tree[si * 2 + 1] + tree[si * 2 + 2];
    }

    // Function to update a range of values in segment
    // tree
    /*  us and eu -> starting and ending indexes of update query
        ue  -> ending index of update query
        diff -> which we need to add in the range us to ue */
    void updateRange(int n, int us, int ue, int diff) {
        updateRangeUtil(0, 0, n - 1, us, ue, diff);
    }

    /*  A recursive function to get the sum of values in given
        range of the array. The following are parameters for
        this function.
        si --> Index of current node in the segment tree.
               Initially 0 is passed as root is always at'
               index 0
        ss & se  --> Starting and ending indexes of the
                     segment represented by current node,
                     i.e., tree[si]
        qs & qe  --> Starting and ending indexes of query
                     range */
    int getSumUtil(int ss, int se, int qs, int qe, int si) {
        // If lazy flag is set for current node of segment tree,
        // then there are some pending updates. So we need to
        // make sure that the pending updates are done before
        // processing the sub sum query
        if (lazy[si] != 0) {
            // Make pending updates to this node. Note that this
            // node represents sum of elements in arr[ss..se] and
            // all these elements must be increased by lazy[si]
            tree[si] += (se - ss + 1) * lazy[si];

            // checking if it is not leaf node because if
            // it is leaf node then we cannot go further
            if (ss != se) {
                // Since we are not yet updating children os si,
                // we need to set lazy values for the children
                lazy[si * 2 + 1] += lazy[si];
                lazy[si * 2 + 2] += lazy[si];
            }

            // unset the lazy value for current node as it has
            // been updated
            lazy[si] = 0;
        }

        // Out of range
        if (ss > se || ss > qe || se < qs)
            return 0;

        // At this point sure, pending lazy updates are done
        // for current node. So we can return value (same as
        // was for query in our previous post)

        // If this segment lies in range
        if (ss >= qs && se <= qe)
            return tree[si];

        // If a part of this segment overlaps with the given
        // range
        int mid = (ss + se) / 2;
        return getSumUtil(ss, mid, qs, qe, 2 * si + 1) +
                getSumUtil(mid + 1, se, qs, qe, 2 * si + 2);
    }

    // Return sum of elements in range from index qs (query
    // start) to qe (query end).  It mainly uses getSumUtil()
    int getSum(int n, int qs, int qe) {
        // Check for erroneous input values
        if (qs < 0 || qe > n - 1 || qs > qe) {

            return 0;
        }

        return getSumUtil(0, n - 1, qs, qe, 0);
    }

    /* A recursive function that constructs Segment Tree for
      array[ss..se]. si is index of current node in segment
      tree st. */
    void constructSTUtil(int arr[], int ss, int se, int si) {
        // out of range as ss can never be greater than se
        if (ss > se)
            return;

        /* If there is one element in array, store it in
         current node of segment tree and return */
        if (ss == se) {
            tree[si] = arr[ss];
            return;
        }

        /* If there are more than one elements, then recur
           for left and right subtrees and store the sum
           of values in this node */
        int mid = (ss + se) / 2;
        constructSTUtil(arr, ss, mid, si * 2 + 1);
        constructSTUtil(arr, mid + 1, se, si * 2 + 2);

        tree[si] = tree[si * 2 + 1] + tree[si * 2 + 2];
    }

    /* Function to construct segment tree from given array.
       This function allocates memory for segment tree and
       calls constructSTUtil() to fill the allocated memory */
    void constructST(int arr[], int n) {
        // Fill the allocated memory st
        constructSTUtil(arr, 0, n - 1, 0);
    }
}

class pair  implements Comparable
{

    int key;
    int value;
    public pair(Object key, Object value) {

        this.key = (int)key;
        this.value=(int)value;
    }

    @Override
    public int compareTo(Object o) {
        pair temp =(pair)o;
        return  key-temp.key;
    }
}
class Graph {


    int n;
    ArrayList<Integer>[] adjList;

    public Graph(int n) {
        this.n = n;
        adjList = new ArrayList[n];
        for (int i = 0; i < n; i++)
            adjList[i] = new ArrayList<>();
    }

}


class InputReader {

    private final InputStream stream;
    private final byte[] buf = new byte[8192];
    private int curChar, snumChars;

    public InputReader(InputStream stream) {
        this.stream = stream;
    }

    public int read() {
        if (snumChars == -1)
            throw new InputMismatchException();
        if (curChar >= snumChars) {
            curChar = 0;
            try {
                snumChars = stream.read(buf);
            } catch (IOException e) {
                throw new InputMismatchException();
            }
            if (snumChars <= 0)
                return -1;
        }
        return buf[curChar++];
    }

    public int nextInt() {
        int c = read();
        while (isSpaceChar(c)) {
            c = read();
        }
        int sgn = 1;
        if (c == '-') {
            sgn = -1;
            c = read();
        }
        int res = 0;
        do {
            res *= 10;
            res += c - '0';
            c = read();
        } while (!isSpaceChar(c));
        return res * sgn;
    }

    public long nextLong() {
        int c = read();
        while (isSpaceChar(c)) {
            c = read();
        }
        int sgn = 1;
        if (c == '-') {
            sgn = -1;
            c = read();
        }
        long res = 0;
        do {
            res *= 10;
            res += c - '0';
            c = read();
        } while (!isSpaceChar(c));
        return res * sgn;
    }

    public int[] nextIntArray(int n) {
        int a[] = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = nextInt();
        }
        return a;
    }

    public String readString() {
        int c = read();
        while (isSpaceChar(c)) {
            c = read();
        }
        StringBuilder res = new StringBuilder();
        do {
            res.appendCodePoint(c);
            c = read();
        } while (!isSpaceChar(c));
        return res.toString();
    }

    public String nextLine() {
        int c = read();
        while (isSpaceChar(c))
            c = read();
        StringBuilder res = new StringBuilder();
        do {
            res.appendCodePoint(c);
            c = read();
        } while (!isEndOfLine(c));
        return res.toString();
    }

    public boolean isSpaceChar(int c) {
        return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
    }

    private boolean isEndOfLine(int c) {
        return c == '\n' || c == '\r' || c == -1;
    }

}

class OutputWriter {
    private final PrintWriter writer;

    public OutputWriter(OutputStream outputStream) {
        writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
    }

    public OutputWriter(Writer writer) {
        this.writer = new PrintWriter(writer);
    }

    public void print(Object... objects) {
        for (int i = 0; i < objects.length; i++) {
            if (i != 0)
                writer.print(' ');
            writer.print(objects[i]);
        }
    }


    public void printLine(Object... objects) {
        print(objects);
        writer.println();
    }

    public void close() {
        writer.close();
    }

    public void flush() {
        writer.flush();
    }

}