#include<iostream>
#include<cstdio>
#include<vector>
#include<cstring>
#include<algorithm>

#define Lens 500010
using namespace std;
struct Node
{
	int key;
	Node *p, *l, *r;
};
int N;
Node *root, *Nil;
char A[20];

void insert(int i)
{
	Node *x = root;
	Node *y = Nil;
	Node *z;
	z = (Node*)malloc(sizeof(Node));
	z->l = z->p = z->r = Nil;
	z->key = i;
	while (x != Nil)
	{
		y = x;
		if (z->key < x->key)
			x = x->l;
		else
			x = x->r;
	}
	if (y == Nil)
		root = z;
	else
	{
		if (z->key < y->key)
		{
			y->l = z;
			z->p = y;
		}
		else
		{
			y->r = z;
			z->p = y;
		}
	}
}

bool Find(int i)
{
	Node *x = root;
	while (x != Nil)
	{
		if (x->key == i)
			return true;
		else if (i < x->key)
			x = x->l;
		else
			x = x->r;
	}
	return false;
}

void Pre(Node *x)
{
	if (x == Nil)
		return;
	cout << ' ' << x->key;
	Pre(x->l);
	Pre(x->r);
}

void In(Node *x)
{
	if (x == Nil)
	{
		return;
	}
	In(x->l);
	cout << ' ' << x->key;
	In(x->r);
}

void Print()
{
	In(root);
	cout << endl;
	Pre(root);
	cout << endl;
}

int main()
{
	int k;
	cin >> N;
	for (int i = 0; i < N; i++)
	{
		cin >> A;
		if (A[0] == 'i')
		{
			cin >> k;
			insert(k);

		}
		else if (A[0] == 'f')
		{
			cin >> k;
			if (Find(k))
				cout << "yes" << endl;
			else
				cout << "no" << endl;
		}
		else if (A[0] == 'p')
			Print();
			
	}
	return 0;
}

### Prompt

Your challenge is to write 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<iostream>
#include<cstdio>
#include<vector>
#include<cstring>
#include<algorithm>

#define Lens 500010
using namespace std;
struct Node
{
	int key;
	Node *p, *l, *r;
};
int N;
Node *root, *Nil;
char A[20];

void insert(int i)
{
	Node *x = root;
	Node *y = Nil;
	Node *z;
	z = (Node*)malloc(sizeof(Node));
	z->l = z->p = z->r = Nil;
	z->key = i;
	while (x != Nil)
	{
		y = x;
		if (z->key < x->key)
			x = x->l;
		else
			x = x->r;
	}
	if (y == Nil)
		root = z;
	else
	{
		if (z->key < y->key)
		{
			y->l = z;
			z->p = y;
		}
		else
		{
			y->r = z;
			z->p = y;
		}
	}
}

bool Find(int i)
{
	Node *x = root;
	while (x != Nil)
	{
		if (x->key == i)
			return true;
		else if (i < x->key)
			x = x->l;
		else
			x = x->r;
	}
	return false;
}

void Pre(Node *x)
{
	if (x == Nil)
		return;
	cout << ' ' << x->key;
	Pre(x->l);
	Pre(x->r);
}

void In(Node *x)
{
	if (x == Nil)
	{
		return;
	}
	In(x->l);
	cout << ' ' << x->key;
	In(x->r);
}

void Print()
{
	In(root);
	cout << endl;
	Pre(root);
	cout << endl;
}

int main()
{
	int k;
	cin >> N;
	for (int i = 0; i < N; i++)
	{
		cin >> A;
		if (A[0] == 'i')
		{
			cin >> k;
			insert(k);

		}
		else if (A[0] == 'f')
		{
			cin >> k;
			if (Find(k))
				cout << "yes" << endl;
			else
				cout << "no" << endl;
		}
		else if (A[0] == 'p')
			Print();
			
	}
	return 0;
}

```