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

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

Node* root;
Node* 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 key)
{
	Node* y = NIL;
	Node* x = root;
	Node* z;

	z = new 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* u)
{
	if (u == NIL)
	{
		return;
	}
	inOrder(u->left);
	cout << " " << u->key;
	inOrder(u->right);
}

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

int main()
{
	int m;
	cin >> m;
	for (int i = 0; i < m; ++i)
	{
		string command;
		cin >> command;
		if (command == "insert")
		{
			int key;
			cin >> key;
			insert(key);
		}
		else if (command == "print")
		{
			inOrder(root);
			cout << endl;
			preOrder(root);
			cout << endl;
		}
		else if (command == "find")
		{
			int key;
			cin >> key;
			Node* t = find(root, key);
			if (t != NIL)
			{
				cout << "yes" << endl;
			}
			else
			{
				cout << "no" << 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 <string>
using namespace std;

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

Node* root;
Node* 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 key)
{
	Node* y = NIL;
	Node* x = root;
	Node* z;

	z = new 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* u)
{
	if (u == NIL)
	{
		return;
	}
	inOrder(u->left);
	cout << " " << u->key;
	inOrder(u->right);
}

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

int main()
{
	int m;
	cin >> m;
	for (int i = 0; i < m; ++i)
	{
		string command;
		cin >> command;
		if (command == "insert")
		{
			int key;
			cin >> key;
			insert(key);
		}
		else if (command == "print")
		{
			inOrder(root);
			cout << endl;
			preOrder(root);
			cout << endl;
		}
		else if (command == "find")
		{
			int key;
			cin >> key;
			Node* t = find(root, key);
			if (t != NIL)
			{
				cout << "yes" << endl;
			}
			else
			{
				cout << "no" << endl;
			}
		}
	}
	return 0;
}

```