Write a program which performs the following operations to a binary search tree $T$ by adding delete operation to B: Binary Search Tree II.

* insert  $k$: Insert a node containing $k$ as key into $T$.
* find $k$: Report whether $T$ has a node containing $k$.
* delete $k$: Delete a node containing $k$.
* print: Print the keys of the binary search tree by inorder tree walk and preorder tree walk respectively.



The operation delete $k$ for deleting a given node $z$ containing key $k$ from $T$ can be implemented by an algorithm which considers the following cases:

1. If $z$ has no children, we modify its parent $z.p$ to replace $z$ with NIL as its child (delete $z$).
2. If $z$ has only a single child, we "splice out" $z$ by making a new link between its child and its parent.
3. If $z$ has two children, we splice out $z$'s successor $y$ and replace $z$'s key with $y$'s key.

Constraints

* The number of operations $\leq 500,000$
* The number of print operations $\leq 10$.
* $-2,000,000,000 \leq key \leq 2,000,000,000$
* The height of the binary tree does not exceed 100 if you employ the above pseudo code.
* The keys in the binary search tree are all different.

Input

In the first line, the number of operations $m$ is given. In the following $m$ lines, operations represented by insert $k$, find $k$, delete $k$ or print are given.

Output

For each find $k$ operation, print "yes" if $T$ has a node containing $k$, "no" if not.

In addition, for each print operation, print a list of keys obtained by inorder tree walk and preorder tree walk in a line respectively. Put a space character before each key

Example

Input

18
insert 8
insert 2
insert 3
insert 7
insert 22
insert 1
find 1
find 2
find 3
find 4
find 5
find 6
find 7
find 8
print
delete 3
delete 7
print


Output

yes
yes
yes
no
no
no
yes
yes
 1 2 3 7 8 22
 8 2 1 3 7 22
 1 2 8 22
 8 2 1 22

import java.util.Scanner;
public class Main {
    static Node nil = new Node(Integer.MIN_VALUE, null, null, null);
    static Node root = nil;
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        for (int i = 0; i < n; i++) {
            String order = sc.next();
            if (order.startsWith("i")) {
                    insert(new Node(sc.nextInt(), nil, nil, nil));
                    
            } else if (order.startsWith("p")){
                printInorder(root);
                System.out.println();
                printPreorder(root);
                System.out.println();
                
            } else if (order.startsWith("f")){
                System.out.println(findNodeWithKey(sc.nextInt()).equals(nil) ? "no" : "yes");
                
            } else {
                int in  = sc.nextInt();
                deleteNodeWithKey(in);
            }
        }
    }
    static void deleteNodeWithKey(int key) {
        Node nd = findNodeWithKey(key);
        Node toDel = nil; // 削除する節点の候補
        if (nd.left.equals(nil) || nd.right.equals(nil)) {
            toDel = nd; // z が子をもたないか、子を1つもつ場合は z の節点そのものを削除
        } else {
            toDel = getSuccessor(nd); // ｚ が子を2つもつ場合は z の次節点
            // 次節点とは、二分探索技における中間順巡回でzの次に訪問される節点
        }

        Node childOfDeled = nil;
        if (!toDel.left.equals(nil)) {
            childOfDeled = toDel.left;
        } else {
            childOfDeled = toDel.right;
        }
        
        if (!childOfDeled.equals(nil)) {
            childOfDeled.parent = toDel.parent;
        }

        if (toDel.parent.equals(nil)) {
            root = childOfDeled;
        } else if (toDel.equals(toDel.parent.left)) {
            toDel.parent.left = childOfDeled;
        } else {
            toDel.parent.right = childOfDeled;
        }

        if (!toDel.equals(nd)) {
            nd.val = toDel.val;
        }
    }
    
    // 次節点の取得
    static Node getSuccessor(Node nd) {
        if (!nd.right.equals(nil)) {
            return getMinimumNode(nd.right);
        }
        
        Node pa = nd.parent;
        while (!pa.equals(nil) && nd.equals(pa.right)) {    // nd.equals(pa.right) で「親paは、「右の子になっているような節点」である」ことを表せる。
            nd = pa;
            pa = pa.parent;
        }
        return pa;
    }
    
    static Node getMinimumNode(Node arg) {
        Node nd = arg;
        while (!nd.left.equals(nil)) {
            nd = nd.left;
        }
        return nd;
    }
    
    static Node findNodeWithKey(int key) {
        Node nd = root;
        while (!nd.equals(nil) && nd.val != key) {
                nd = (nd.val > key) ? nd.left : nd.right;
        }
        return nd;
    }
    
    static void insert(Node z) {
        if (root.equals(nil)) {
            root = z;
            return;
        }
        
        Node p = root; // zの親
        while (true) {
            if (p.val > z.val) {
                if (p.left.equals(nil)) {
                    p.left = z;
                    break;
                } else {
                    p = p.left;
                }
            } else {
                if (p.right.equals(nil)) {
                    p.right = z;
                    break;
                } else {
                    p = p.right;
                }
            }
        }
        z.parent = p;
    }
    
    static void printInorder(Node arg) {
        Node nd = arg;
        if(!nd.equals(nil)) {
            printInorder(nd.left);
            System.out.print(" " + nd.val);
            printInorder(nd.right);
        }
    }
    
    static void printPreorder(Node arg) {
        Node nd = arg;
        if(!nd.equals(nil)) {
            System.out.print(" " + nd.val);
            printPreorder(nd.left);
            printPreorder(nd.right);
        }
    }
    
    static class Node {
        int val;
        Node parent;
        Node left;
        Node right;
        
        Node (int val, Node parent, Node left, Node right) {
            this.val = val;
            this.parent = parent;
            this.left = left;
            this.right = right;
        }
        Node (int val) {
            this.val = val;
            this.parent = nil;
            this.left = nil;
            this.right = nil;
        }
    }
}

