#include<iostream>
#include<stdlib.h>
using namespace std;

struct node{
  int key;
  node *right, *left, *parent;
};

node *root, *nil;

node *find(node *u, int k){
  while(u != nil && k != u->key){
    if(k < u->key) u = u->left;

    else u = u->right;
    
  }

  return u;
  
}

void insert(int k){
  node *y = nil;
  node *x = root;
  node *z;

  z = (node *)malloc(sizeof(node));
  z->key = k;
  z->left = nil;
  z->right = nil;

  while(x != nil){
    y = x;

    if(z->key < x->key) x = x->left;
    else x = x->right;

  }

    z->parent = y;

    if(y == nil) root = z;

    else{
      if(z->key < y->key) y->left = z;

      else y->right = z;

  }  
}

void inorder(node *a){
  if(a == nil) return;

  inorder(a->left);

  cout << " " << a->key;

  inorder(a->right);
}

void preorder(node *a){
  if(a == nil) return;

  cout << " " << a->key;
  
  preorder(a->left);
  preorder(a->right);
  
}

int main()
{
  int n, x;
  string s;

  cin >> n;

  for(int i = 0; i < n; i++){
    cin >> s;

    if(s == "find"){
      cin >> x;
      node *t = find(root, x);

      if(t != nil) cout << "yes" << endl;
      else cout << "no" << endl;
      
    }

    else if(s == "insert"){
      cin >> x;
      insert(x);
     
    }

    else if(s == "print"){
      inorder(root);
      cout << endl;

      preorder(root);
      cout << endl;
    }
  }

  return 0;
  
}


### Prompt

Construct a cpp code solution to the problem outlined:

Write a program which performs the following operations to a binary search tree $T$ by adding the find operation to A: Binary Search Tree I.

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

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 $k$ 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

10
insert 30
insert 88
insert 12
insert 1
insert 20
find 12
insert 17
insert 25
find 16
print


Output

yes
no
 1 12 17 20 25 30 88
 30 12 1 20 17 25 88

### Response

```cpp
#include<iostream>
#include<stdlib.h>
using namespace std;

struct node{
  int key;
  node *right, *left, *parent;
};

node *root, *nil;

node *find(node *u, int k){
  while(u != nil && k != u->key){
    if(k < u->key) u = u->left;

    else u = u->right;
    
  }

  return u;
  
}

void insert(int k){
  node *y = nil;
  node *x = root;
  node *z;

  z = (node *)malloc(sizeof(node));
  z->key = k;
  z->left = nil;
  z->right = nil;

  while(x != nil){
    y = x;

    if(z->key < x->key) x = x->left;
    else x = x->right;

  }

    z->parent = y;

    if(y == nil) root = z;

    else{
      if(z->key < y->key) y->left = z;

      else y->right = z;

  }  
}

void inorder(node *a){
  if(a == nil) return;

  inorder(a->left);

  cout << " " << a->key;

  inorder(a->right);
}

void preorder(node *a){
  if(a == nil) return;

  cout << " " << a->key;
  
  preorder(a->left);
  preorder(a->right);
  
}

int main()
{
  int n, x;
  string s;

  cin >> n;

  for(int i = 0; i < n; i++){
    cin >> s;

    if(s == "find"){
      cin >> x;
      node *t = find(root, x);

      if(t != nil) cout << "yes" << endl;
      else cout << "no" << endl;
      
    }

    else if(s == "insert"){
      cin >> x;
      insert(x);
     
    }

    else if(s == "print"){
      inorder(root);
      cout << endl;

      preorder(root);
      cout << endl;
    }
  }

  return 0;
  
}


```