#include <cstdio>
#include <iostream>
#include <string>
#include <cstdlib>

using namespace std;

struct Node{
  int key;
  Node *parent,*left,*right;
};
Node *root,*NIL;

void find(int k){
  Node* x=root;
  while(x!=NIL){
    if(k<x->key){
      x=x->left;
    }else if(k==x->key){
      cout << "yes\n";
      return;
    }else{
      x=x->right;
    }
  }
  cout << "no\n";
}

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(y==NIL){
      root=z;
    }else{
      if(z->key<y->key) y->left=z;
      else y->right=z;
    }
  }
}

void inorder(Node* u){
  if(u==NIL) return;
  inorder(u->left);
  printf(" %d",u->key);
  inorder(u->right);
}

void preorder(Node* u){
  if(u==NIL) return;
  printf(" %d",u->key);
  preorder(u->left);
  preorder(u->right);
}

int main(){
  int n;
  scanf("%d",&n);
  for(int i=0;i<n;i++){
    string com;
    cin >> com;
    if(com=="print"){
      inorder(root);
      cout << endl;
      preorder(root);
      cout << endl;
    }
    else if(com=="insert"){
      int z;
      cin >> z;
      insert(z);
    }
    else if(com=="find"){
      int z;
      cin >> z;
      find(z);
    }
  }
}
### Prompt

Generate a Cpp solution to the following problem:

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 <cstdio>
#include <iostream>
#include <string>
#include <cstdlib>

using namespace std;

struct Node{
  int key;
  Node *parent,*left,*right;
};
Node *root,*NIL;

void find(int k){
  Node* x=root;
  while(x!=NIL){
    if(k<x->key){
      x=x->left;
    }else if(k==x->key){
      cout << "yes\n";
      return;
    }else{
      x=x->right;
    }
  }
  cout << "no\n";
}

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(y==NIL){
      root=z;
    }else{
      if(z->key<y->key) y->left=z;
      else y->right=z;
    }
  }
}

void inorder(Node* u){
  if(u==NIL) return;
  inorder(u->left);
  printf(" %d",u->key);
  inorder(u->right);
}

void preorder(Node* u){
  if(u==NIL) return;
  printf(" %d",u->key);
  preorder(u->left);
  preorder(u->right);
}

int main(){
  int n;
  scanf("%d",&n);
  for(int i=0;i<n;i++){
    string com;
    cin >> com;
    if(com=="print"){
      inorder(root);
      cout << endl;
      preorder(root);
      cout << endl;
    }
    else if(com=="insert"){
      int z;
      cin >> z;
      insert(z);
    }
    else if(com=="find"){
      int z;
      cin >> z;
      find(z);
    }
  }
}
```