#include <cstdio>

struct node {
  int key;
  node *left, *right;
  node(int key) : key(key), left(nullptr), right(nullptr) {}
  ~node() {
    delete left;
    delete right;
  }
};

enum order {
  pre,
  in
};

void rec(node *t, order o) {
  if (!t)
    return;
  if (o == pre)
    printf(" %d", t->key);
  rec(t->left, o);
  if (o == in)
    printf(" %d", t->key);
  rec(t->right, o);
}

int main() {
  node *T(nullptr);
  size_t m;
  scanf("%zu\n", &m);
  for (size_t i = 0; i < m; i++) {
    char s[8];
    scanf("%s", s);
    switch (s[0]) {
      case 'i':
        {
          int k;
          scanf("%d", &k);
          node *x = T, *y = nullptr, *z = new node(k);
          while (x) {
            y = x;
            x = z->key < y->key ? y->left : y->right;
          }
          (y ? z->key < y->key ? y->left : y->right : T) = z;
        }
        break;
      case 'f':
        {
          int k;
          scanf("%d", &k);
          bool found = false;
          for (auto x = T; x; x = k < x->key ? x->left : x->right)
            if (x->key == k) {
              found = true;
              break;
            }
          puts(found ? "yes" : "no");
        }
        break;
      default:
        rec(T, in);
        puts("");
        rec(T, pre);
        puts("");
        break;
    }
  }
  delete T;
  return 0;
}
### Prompt

Create a solution in CPP for 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>

struct node {
  int key;
  node *left, *right;
  node(int key) : key(key), left(nullptr), right(nullptr) {}
  ~node() {
    delete left;
    delete right;
  }
};

enum order {
  pre,
  in
};

void rec(node *t, order o) {
  if (!t)
    return;
  if (o == pre)
    printf(" %d", t->key);
  rec(t->left, o);
  if (o == in)
    printf(" %d", t->key);
  rec(t->right, o);
}

int main() {
  node *T(nullptr);
  size_t m;
  scanf("%zu\n", &m);
  for (size_t i = 0; i < m; i++) {
    char s[8];
    scanf("%s", s);
    switch (s[0]) {
      case 'i':
        {
          int k;
          scanf("%d", &k);
          node *x = T, *y = nullptr, *z = new node(k);
          while (x) {
            y = x;
            x = z->key < y->key ? y->left : y->right;
          }
          (y ? z->key < y->key ? y->left : y->right : T) = z;
        }
        break;
      case 'f':
        {
          int k;
          scanf("%d", &k);
          bool found = false;
          for (auto x = T; x; x = k < x->key ? x->left : x->right)
            if (x->key == k) {
              found = true;
              break;
            }
          puts(found ? "yes" : "no");
        }
        break;
      default:
        rec(T, in);
        puts("");
        rec(T, pre);
        puts("");
        break;
    }
  }
  delete T;
  return 0;
}
```