#include<iostream>
#include<cstdio>
#include<string>
#include<cstdlib>
using namespace std;
struct node{
	int key;
	node *right, *left, *p;
};

node *root, *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 k)
{
	 node *y=NIL;
	 node *x=root;
	 node *z;
	 z=(node *)malloc(sizeof(node));
	 z->key=k;
	 z->left=NIL;
	 z->right=NIL;
	 while(x!=NIL)
	 {
	 	y=x;
	 	if(z->key < x->key)
	 	{
	 		x=x->left;
		 }
		 else {
		 	x=x->right;
		 }
	 }
	 z->p=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);
	printf(" %d",u->key);
	inorder(u->right);
}

void preorder(node *u)
{
	if(u==NIL)return;
	printf(" %d",u->key);
	preorder(u->left);
	preorder(u->right);
}

int main()
{
	int n,i,x;
	string com;
	scanf("%d",&n);
	for(i=0;i<n;i++)
	{
		cin>>com;
		if(com[0]=='f')
		{
			scanf("%d",&x);
			node *t=find(root,x);
			if(t!=NIL)printf("yes\n");
			else printf("no\n");
		}
		else if(com=="insert")
		{
		scanf("%d",&x);
		insert(x);
		}
		else if(com=="print")
		{
			inorder(root);
			printf("\n");
			preorder(root);
			printf("\n");
		}
	}
	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<iostream>
#include<cstdio>
#include<string>
#include<cstdlib>
using namespace std;
struct node{
	int key;
	node *right, *left, *p;
};

node *root, *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 k)
{
	 node *y=NIL;
	 node *x=root;
	 node *z;
	 z=(node *)malloc(sizeof(node));
	 z->key=k;
	 z->left=NIL;
	 z->right=NIL;
	 while(x!=NIL)
	 {
	 	y=x;
	 	if(z->key < x->key)
	 	{
	 		x=x->left;
		 }
		 else {
		 	x=x->right;
		 }
	 }
	 z->p=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);
	printf(" %d",u->key);
	inorder(u->right);
}

void preorder(node *u)
{
	if(u==NIL)return;
	printf(" %d",u->key);
	preorder(u->left);
	preorder(u->right);
}

int main()
{
	int n,i,x;
	string com;
	scanf("%d",&n);
	for(i=0;i<n;i++)
	{
		cin>>com;
		if(com[0]=='f')
		{
			scanf("%d",&x);
			node *t=find(root,x);
			if(t!=NIL)printf("yes\n");
			else printf("no\n");
		}
		else if(com=="insert")
		{
		scanf("%d",&x);
		insert(x);
		}
		else if(com=="print")
		{
			inorder(root);
			printf("\n");
			preorder(root);
			printf("\n");
		}
	}
	return 0;
}


```