#include<bits/stdc++.h>
using namespace std;
#define int long long

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

void insert(int k){
  node *p;
  p = (node *)malloc(sizeof(node));
  p->key = k;
  p->left = nil;
  p->right = nil;

  node *v = root,*u = nil;
  //if(v == nil) return;
  while(v != nil){
    u = v;
    if(p->key < v->key) v = v->left;
    else v = v->right;
  }
  p->parent = u;
  if(u == nil) root = p;
  else {
    if(p->key < u->key) u->left = p;
    else u->right = p;
  }
}

bool find(int k){
  node *p;
  p = root;
  while(p != nil){
    if(p->key == k) return 1;
    if(p->key < k) p = p->right;
    else p = p->left;
  }
  return 0;
}

void preorder(node *u){
  if(u == nil) return;
  cout << " " << u->key;
  preorder(u->left);
  preorder(u->right);
}

void inorder(node *u){
  if(u == nil) return;
  inorder(u->left);
  cout << " " << u->key;
  inorder(u->right);
}

signed main(){
  cin.tie(0);
  ios::sync_with_stdio(false);

  int m; cin >> m;
  for(int i = 0; i < m; i++){
    string s; cin >> s;
    if(s[0] == 'p'){
      inorder(root);
      cout << '\n';
      preorder(root);
      cout << '\n';
      continue;
    }
    else if(s[0] == 'f'){
      int k; cin >> k;
      if(find(k)) cout << "yes" << endl;
      else cout << "no" << endl;
      continue;
    }
    int k; cin >> k;
    insert(k);
  }

  return 0;
}


### 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<bits/stdc++.h>
using namespace std;
#define int long long

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

void insert(int k){
  node *p;
  p = (node *)malloc(sizeof(node));
  p->key = k;
  p->left = nil;
  p->right = nil;

  node *v = root,*u = nil;
  //if(v == nil) return;
  while(v != nil){
    u = v;
    if(p->key < v->key) v = v->left;
    else v = v->right;
  }
  p->parent = u;
  if(u == nil) root = p;
  else {
    if(p->key < u->key) u->left = p;
    else u->right = p;
  }
}

bool find(int k){
  node *p;
  p = root;
  while(p != nil){
    if(p->key == k) return 1;
    if(p->key < k) p = p->right;
    else p = p->left;
  }
  return 0;
}

void preorder(node *u){
  if(u == nil) return;
  cout << " " << u->key;
  preorder(u->left);
  preorder(u->right);
}

void inorder(node *u){
  if(u == nil) return;
  inorder(u->left);
  cout << " " << u->key;
  inorder(u->right);
}

signed main(){
  cin.tie(0);
  ios::sync_with_stdio(false);

  int m; cin >> m;
  for(int i = 0; i < m; i++){
    string s; cin >> s;
    if(s[0] == 'p'){
      inorder(root);
      cout << '\n';
      preorder(root);
      cout << '\n';
      continue;
    }
    else if(s[0] == 'f'){
      int k; cin >> k;
      if(find(k)) cout << "yes" << endl;
      else cout << "no" << endl;
      continue;
    }
    int k; cin >> k;
    insert(k);
  }

  return 0;
}


```