#include <iostream>
using namespace std;
struct node{
	int key;
	node *left = NULL;
	node *right = NULL;
	node(int k, node*l = NULL, node*r = NULL) : key(k), left(l), right(r) {}
};
node *Tree;

void insert(node z)
{
	node *y = NULL;
	node *x = Tree;
	while(x != NULL){
		y = x;
		if(z.key < x->key)
			x = x->left;
		else
			x = x->right;
	}
	if(y == NULL)
		Tree = new node(z.key);
	else if(z.key < y->key)
		y->left = new node(z.key);
	else
		y->right = new node(z.key);
}

void midprint(node *root)
{
	if(root == NULL) return;
	midprint(root->left);
	cout << " " << root->key;
	midprint(root->right);
}

void preprint(node *root)
{
	if(root == NULL) return;
	cout << " " << root->key ;
	preprint(root->left);
	preprint(root->right);
}

void find(int k)
{
	node *x = Tree;
	while(x != NULL){
		if(k == x->key){
			cout << "yes" << endl;
			return ;
		}
		if(k < x->key)
			x = x->left;
		else
			x = x->right;
	}
	cout << "no" << endl;
	return;
}


void print(node *root)
{
	midprint(root);
	cout << endl;
	preprint(root);
	cout << endl;
}

int main()
{
	int n;
	cin >> n;
	while(n--){
		string s;
		cin >> s;
		if(s[0] == 'i'){
			int k;
			cin >> k;
			node leaf(k);
			insert(leaf);
		}
		else if(s[0] == 'p')
			print(Tree);
		else if(s[0] == 'f'){
			int k;
			cin >> k;
			find(k);
		}
	}
	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>
using namespace std;
struct node{
	int key;
	node *left = NULL;
	node *right = NULL;
	node(int k, node*l = NULL, node*r = NULL) : key(k), left(l), right(r) {}
};
node *Tree;

void insert(node z)
{
	node *y = NULL;
	node *x = Tree;
	while(x != NULL){
		y = x;
		if(z.key < x->key)
			x = x->left;
		else
			x = x->right;
	}
	if(y == NULL)
		Tree = new node(z.key);
	else if(z.key < y->key)
		y->left = new node(z.key);
	else
		y->right = new node(z.key);
}

void midprint(node *root)
{
	if(root == NULL) return;
	midprint(root->left);
	cout << " " << root->key;
	midprint(root->right);
}

void preprint(node *root)
{
	if(root == NULL) return;
	cout << " " << root->key ;
	preprint(root->left);
	preprint(root->right);
}

void find(int k)
{
	node *x = Tree;
	while(x != NULL){
		if(k == x->key){
			cout << "yes" << endl;
			return ;
		}
		if(k < x->key)
			x = x->left;
		else
			x = x->right;
	}
	cout << "no" << endl;
	return;
}


void print(node *root)
{
	midprint(root);
	cout << endl;
	preprint(root);
	cout << endl;
}

int main()
{
	int n;
	cin >> n;
	while(n--){
		string s;
		cin >> s;
		if(s[0] == 'i'){
			int k;
			cin >> k;
			node leaf(k);
			insert(leaf);
		}
		else if(s[0] == 'p')
			print(Tree);
		else if(s[0] == 'f'){
			int k;
			cin >> k;
			find(k);
		}
	}
	return 0;
}
```