#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MOD = 1e9+7,MAX = 1e5;

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

Node *root;

void insert(int k)
{
	Node *y = nullptr;
	Node *x = root;
	Node *z ;

	z = new Node;
	z->key = k;
	z->left = nullptr;
	z->right = nullptr;

	while(x!=nullptr)
	{
		y=x;
		if(z->key<x->key)
		{
			x = x->left;
		}
		else
		{
			x = x ->right;
		}
	}
	
	z -> parent = y;
	if(y == nullptr)
	{
		root = z;
	}
	else
	{
		if(z->key < y->key)
		{
			y->left = z;
		}
		else
		{
			y -> right = z;
		}
	}
}

void inorder(Node *u)
{
	if(u == nullptr)return;
	inorder(u->left);
	cout<<' '<<u->key;
	inorder(u->right);
}

void preorder(Node *u)
{
	if(u==nullptr)return;
	cout<<' '<<u->key;
	preorder(u->left);
	preorder(u->right);
}

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

int main()
{
	int n,x;
	cin >> n;
	string s;
	for(int i =0;i<n;i++)
	{
		cin >> s;
		if(s=="insert")
		{
			cin >> x;
			insert (x);
		}
		else if(s=="find")
		{
			cin >> x;
			if(find(root,x)!=nullptr)cout<<"yes"<<endl;
			else cout<<"no"<<endl;
		}
		else
		{
			inorder(root);cout<<endl;
			preorder(root);cout<<endl;
		}
	}
	return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
typedef long long ll;
const int MOD = 1e9+7,MAX = 1e5;

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

Node *root;

void insert(int k)
{
	Node *y = nullptr;
	Node *x = root;
	Node *z ;

	z = new Node;
	z->key = k;
	z->left = nullptr;
	z->right = nullptr;

	while(x!=nullptr)
	{
		y=x;
		if(z->key<x->key)
		{
			x = x->left;
		}
		else
		{
			x = x ->right;
		}
	}
	
	z -> parent = y;
	if(y == nullptr)
	{
		root = z;
	}
	else
	{
		if(z->key < y->key)
		{
			y->left = z;
		}
		else
		{
			y -> right = z;
		}
	}
}

void inorder(Node *u)
{
	if(u == nullptr)return;
	inorder(u->left);
	cout<<' '<<u->key;
	inorder(u->right);
}

void preorder(Node *u)
{
	if(u==nullptr)return;
	cout<<' '<<u->key;
	preorder(u->left);
	preorder(u->right);
}

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

int main()
{
	int n,x;
	cin >> n;
	string s;
	for(int i =0;i<n;i++)
	{
		cin >> s;
		if(s=="insert")
		{
			cin >> x;
			insert (x);
		}
		else if(s=="find")
		{
			cin >> x;
			if(find(root,x)!=nullptr)cout<<"yes"<<endl;
			else cout<<"no"<<endl;
		}
		else
		{
			inorder(root);cout<<endl;
			preorder(root);cout<<endl;
		}
	}
	return 0;
}

```