Problem:
Treap

A binary search tree can be unbalanced depending on features of data. For example, if we insert $n$ elements into a binary search tree in ascending order, the tree become a list, leading to long search times. One of strategies is to randomly shuffle the elements to be inserted. However, we should consider to maintain the balanced binary tree where different operations can be performed one by one depending on requirement.

We can maintain the balanced binary search tree by assigning a priority randomly selected to each node and by ordering nodes based on the following properties. Here, we assume that all priorities are distinct and also that all keys are distinct.

binary-search-tree property. If $v$ is a left child of $u$, then $v.key < u.key$ and if $v$ is a right child of $u$, then $u.key < v.key$

heap property. If $v$ is a child of $u$, then $v.priority < u.priority$

This combination of properties is why the tree is called Treap (tree + heap).

An example of Treap is shown in the following figure.

Insert

To insert a new element into a Treap, first of all, insert a node which a randomly selected priority value is assigned in the same way for ordinal binary search tree. For example, the following figure shows the Treap after a node with key = 6 and priority = 90 is inserted.

It is clear that this Treap violates the heap property, so we need to modify the structure of the tree by rotate operations. The rotate operation is to change parent-child relation while maintaing the binary-search-tree property.

The rotate operations can be implemented as follows.

rightRotate(Node t)
Node s = t.left
t.left = s.right
s.right = t
return s // the new root of subtree

leftRotate(Node t)
Node s = t.right
t.right = s.left
s.left = t
return s // the new root of subtree

The following figure shows processes of the rotate operations after the insert operation to maintain the properties.

The insert operation with rotate operations can be implemented as follows.

insert(Node t, int key, int priority) // search the corresponding place recursively
if t == NIL
return Node(key, priority) // create a new node when you reach a leaf
if key == t.key
return t // ignore duplicated keys

if key < t.key // move to the left child
t.left = insert(t.left, key, priority) // update the pointer to the left child
if t.priority < t.left.priority // rotate right if the left child has higher priority
t = rightRotate(t)
else // move to the right child
t.right = insert(t.right, key, priority) // update the pointer to the right child
if t.priority < t.right.priority // rotate left if the right child has higher priority
t = leftRotate(t)

return t

Delete

To delete a node from the Treap, first of all, the target node should be moved until it becomes a leaf by rotate operations. Then, you can remove the node (the leaf). These processes can be implemented as follows.

delete(Node t, int key) // seach the target recursively
if t == NIL
return NIL
if key < t.key // search the target recursively
t.left = delete(t.left, key)
else if key > t.key
t.right = delete(t.right, key)
else
return _delete(t, key)
return t

_delete(Node t, int key) // if t is the target node
if t.left == NIL && t.right == NIL // if t is a leaf
return NIL
else if t.left == NIL // if t has only the right child, then perform left rotate
t = leftRotate(t)
else if t.right == NIL // if t has only the left child, then perform right rotate
t = rightRotate(t)
else // if t has both the left and right child
if t.left.priority > t.right.priority // pull up the child with higher priority
t = rightRotate(t)
else
t = leftRotate(t)
return delete(t, key)

Write a program which performs the following operations to a Treap $T$ based on the above described algorithm.

insert ($k$, $p$): Insert a node containing $k$ as key and $p$ as priority to $T$.

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.

Input

In the first line, the number of operations $m$ is given. In the following $m$ lines, operations represented by insert $k \; p$, 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.

Constraints

The number of operations $ \leq 200,000$

$0 \leq k, p \leq 2,000,000,000$

The height of the binary tree does not exceed 50 if you employ the above algorithm

The keys in the binary search tree are all different.

The priorities in the binary search tree are all different.

The size of output $\leq 10$ MB

Sample Input 1

16
insert 35 99
insert 3 80
insert 1 53
insert 14 25
insert 80 76
insert 42 3
insert 86 47
insert 21 12
insert 7 10
insert 6 90
print
find 21
find 22
delete 35
delete 99
print

Sample Output 1

1 3 6 7 14 21 35 42 80 86
35 6 3 1 14 7 21 80 42 86
yes
no
1 3 6 7 14 21 42 80 86
6 3 1 80 14 7 21 42 86

Reference

Introduction to Algorithms, Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, and Clifford Stein. The MIT Press.

Randomized Search Trees, R. Seidel, C.R. Aragon.

Solution in Ruby (judge verdict: Accepted)

class Node
  def initialize(key, priority)
    @key = key
    @priority = priority
    @parent = nil
    @left = nil
    @right = nil
  end
  attr_accessor :key, :priority, :parent, :left, :right
end

def right_rotate(node_t)
  node_s = node_t.left
  node_t.left = node_s.right
  node_s.right = node_t
  return node_s
end

def left_rotate(node_t)
  node_s = node_t.right
  node_t.right = node_s.left
  node_s.left = node_t
  return node_s
end

def insert(node_t, key, priority)
  if node_t == nil
    return Node.new(key,priority)
  end
  if key == node_t.key
    return node_t
  end

  if key < node_t.key
    node_t.left = insert(node_t.left, key, priority)
    if node_t.priority < node_t.left.priority
      node_t = right_rotate(node_t)
    end
  else
    node_t.right = insert(node_t.right, key, priority)
    if node_t.priority < node_t.right.priority
      node_t = left_rotate(node_t)
    end
  end
  return node_t
end

def find?(key)
  node_x = @root
  loop do
    if node_x == nil
      return false
    elsif node_x.key == key
      return node_x
    elsif key < node_x.key
      node_x = node_x.left
    else
      node_x = node_x.right
    end
  end
end

def delete(node_t, key)
  if node_t == nil
    return nil
  end
  if key < node_t.key
    node_t.left = delete(node_t.left, key)
  elsif key > node_t.key
    node_t.right = delete(node_t.right, key)
  else
    return _delete(node_t, key)
  end
  return node_t
end

def _delete(node_t, key)
  if node_t.left == nil && node_t.right ==nil
    return nil
  elsif node_t.left == nil
    node_t = left_rotate(node_t)
  elsif node_t.right == nil
    node_t = right_rotate(node_t)
  else
    if node_t.left.priority > node_t.right.priority
      node_t = right_rotate(node_t)
    else
      node_t = left_rotate(node_t)
    end
  end
  return delete(node_t, key)
end


def inorder_tree_walk(node)
  if node != nil
    inorder_tree_walk(node.left)
    print " "
    print node.key
    inorder_tree_walk(node.right)
  end
end

def preorder_tree_walk(node)
  if node != nil
    print " "
    print node.key
    preorder_tree_walk(node.left)
    preorder_tree_walk(node.right)
  end
end

def print_tree
  inorder_tree_walk(@root)
  puts
  preorder_tree_walk(@root)
  puts
end


@root = nil

m = gets.to_i
m.times do |t|
  input = gets.split(" ")
  command = input.shift
  input.map!{|x| x.to_i}
  if command == "insert"
    @root = insert(@root, input[0], input[1])
  elsif command == "find"
    if find?(input[0])
      puts "yes"
    else
      puts "no"
    end
  elsif command == "delete"
    @root = delete(@root, input[0])
  else
    print_tree
  end
end