#include <iostream>
#include <vector>
#include <string>
#include <cstdlib>

using namespace std;

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

node *root,*nil,*z;

void preorder(node *id){
	if(id == nil)return;
	cout << " " << id->key;
	preorder(id->left);
	preorder(id->right);
}

void inorder(node *id){
	if(id == nil)return;
	inorder(id->left);
	cout << " " << id->key;
	inorder(id->right);
}

void print(){
	inorder(root);
	cout << endl;
	preorder(root);
	cout << endl;
}

void insert(int k){
	node *p = nil;
	node *r = root;
	node *z = (node *)malloc(sizeof(node));
	z->key = k;
	z->left = nil;
	z->right = nil;
	while(r != nil){
		p = r;
		if(z->key < r->key){
			r = r->left;
		}else{
			r = r->right;
		}
	}
	z->parent = p;
	if(p == nil){
		root = z;
	}else if(z->key < p->key){
		p->left = z;
	}else{
		p->right = z;
	}
}

node* find(int k,node *i){
	while(i != nil && i->key != k){
		if(i->key > k){
			i = i->left;
		}else{
			i = i->right;
		}
	}
	return i;
}

int main(void){
	int n;
	cin >> n;
	for(int i=0;i<n;i++){
		string order;
		int key;
		cin >> order;
		if(order == "print"){
			print();
		}else if(order == "insert"){
			cin >> key;
			insert(key);
		}else if(order == "find"){
			cin >> key;
			node* res = find(key,root);
			if(res != nil)cout << "yes" << endl;
			else cout << "no" << endl;
		}
	}
}

### Prompt

Please formulate 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 <vector>
#include <string>
#include <cstdlib>

using namespace std;

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

node *root,*nil,*z;

void preorder(node *id){
	if(id == nil)return;
	cout << " " << id->key;
	preorder(id->left);
	preorder(id->right);
}

void inorder(node *id){
	if(id == nil)return;
	inorder(id->left);
	cout << " " << id->key;
	inorder(id->right);
}

void print(){
	inorder(root);
	cout << endl;
	preorder(root);
	cout << endl;
}

void insert(int k){
	node *p = nil;
	node *r = root;
	node *z = (node *)malloc(sizeof(node));
	z->key = k;
	z->left = nil;
	z->right = nil;
	while(r != nil){
		p = r;
		if(z->key < r->key){
			r = r->left;
		}else{
			r = r->right;
		}
	}
	z->parent = p;
	if(p == nil){
		root = z;
	}else if(z->key < p->key){
		p->left = z;
	}else{
		p->right = z;
	}
}

node* find(int k,node *i){
	while(i != nil && i->key != k){
		if(i->key > k){
			i = i->left;
		}else{
			i = i->right;
		}
	}
	return i;
}

int main(void){
	int n;
	cin >> n;
	for(int i=0;i<n;i++){
		string order;
		int key;
		cin >> order;
		if(order == "print"){
			print();
		}else if(order == "insert"){
			cin >> key;
			insert(key);
		}else if(order == "find"){
			cin >> key;
			node* res = find(key,root);
			if(res != nil)cout << "yes" << endl;
			else cout << "no" << endl;
		}
	}
}

```