#include <iostream>
#include <cstdlib>
#include <string>
using namespace std;

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

Node *root, *NIL;

Node *find(Node *root, int key)
{
	while (root != NIL && key != root->key)
	{
		if (key < root->key) root = root->left;
		else root = root->right;
	}
	return root;
}

void insert(int key)
{
	Node *x = root;
	Node *y = NIL;
	Node *z;

        z = (Node *)malloc(sizeof(Node));
	z->key = key;
	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 *root)
{
	if (root == NIL) return;
	inorder(root->left);
	cout << " " << root->key;
	inorder(root->right);
}

void preorder(Node *root)
{
	if (root == NIL) return;
	cout << " " << root->key;
	preorder(root->left);
	preorder(root->right);
}

int main()
{
	int n, key;
	string str;

	cin >> n;
	for (int i = 0; i < n; i++)
	{
		cin >> str;
		if (str == "find")
		{
			cin >> key;
			Node *t = find(root, key);
			if (t != NIL) cout << "yes" << endl;
			else cout << "no" << endl;
		}
		else if (str == "insert")
		{
			cin >> key;
			insert(key);
		}
		else if (str == "print")
		{
			inorder(root);
			cout << endl;
			preorder(root);
			cout << endl;
		}
	}

	return 0;
}
### Prompt

Please create a solution in Cpp 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 <iostream>
#include <cstdlib>
#include <string>
using namespace std;

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

Node *root, *NIL;

Node *find(Node *root, int key)
{
	while (root != NIL && key != root->key)
	{
		if (key < root->key) root = root->left;
		else root = root->right;
	}
	return root;
}

void insert(int key)
{
	Node *x = root;
	Node *y = NIL;
	Node *z;

        z = (Node *)malloc(sizeof(Node));
	z->key = key;
	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 *root)
{
	if (root == NIL) return;
	inorder(root->left);
	cout << " " << root->key;
	inorder(root->right);
}

void preorder(Node *root)
{
	if (root == NIL) return;
	cout << " " << root->key;
	preorder(root->left);
	preorder(root->right);
}

int main()
{
	int n, key;
	string str;

	cin >> n;
	for (int i = 0; i < n; i++)
	{
		cin >> str;
		if (str == "find")
		{
			cin >> key;
			Node *t = find(root, key);
			if (t != NIL) cout << "yes" << endl;
			else cout << "no" << endl;
		}
		else if (str == "insert")
		{
			cin >> key;
			insert(key);
		}
		else if (str == "print")
		{
			inorder(root);
			cout << endl;
			preorder(root);
			cout << endl;
		}
	}

	return 0;
}
```