#include<iostream>
#include<cstdio>
#include<string>
#include<vector>
#include<algorithm>
#define MAX 100
using namespace std;
vector<int> zhongxu,qianxu;
struct node{int key;node *p,*l,*r;};
node *root,*NIL;

void insert(int num){
	node *y=NIL;node *x=root;node *z;
	z=(node *)malloc(sizeof(node ));
	while(x!=NIL){
		y=x;
		if(num<x->key){x=(*x).l;}
		else {x=x->r;}
	}
	z->key=num;z->l=NIL;z->r=NIL;z->p=y;
	if(y==NIL){
		root=z;
	}else if(num>y->key){
		y->r=z;
	}else{y->l=z;}
}
void zhong(node *u){
	if(u->l!=NIL) zhong(u->l);
	cout<<" "<<u->key;
	if(u->r!=NIL) zhong(u->r);
	return ;
}
void qian(node *u){
	cout<<" "<<u->key;
	if(u->l!=NIL) qian(u->l);
	if(u->r!=NIL) qian(u->r);
	return;
}
void print(node *u){
	zhong(u);cout<<endl;
	qian(u);cout<<endl;return;
}
node* find(int num){
	node *p=root;
	 while(p!=NIL&&p->key!=num){
	 	if(num<p->key) p=p->l;
	 	else if(num>p->key) p=p->r;
	 }
	 return p;
}
int main(){
	int cnt;int num;
	scanf("%d",&cnt);
	char order[10];
	vector<int> ::iterator it;
	for(int i=0;i<cnt;i++){
		scanf("%s",&order);
		if(order[0]=='i') {
		scanf("%d",&num);insert(num);
		}
		else if(order[0]=='p') {print(root);
		}else{scanf("%d",&num);
		if(find(num)!=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<cstdio>
#include<string>
#include<vector>
#include<algorithm>
#define MAX 100
using namespace std;
vector<int> zhongxu,qianxu;
struct node{int key;node *p,*l,*r;};
node *root,*NIL;

void insert(int num){
	node *y=NIL;node *x=root;node *z;
	z=(node *)malloc(sizeof(node ));
	while(x!=NIL){
		y=x;
		if(num<x->key){x=(*x).l;}
		else {x=x->r;}
	}
	z->key=num;z->l=NIL;z->r=NIL;z->p=y;
	if(y==NIL){
		root=z;
	}else if(num>y->key){
		y->r=z;
	}else{y->l=z;}
}
void zhong(node *u){
	if(u->l!=NIL) zhong(u->l);
	cout<<" "<<u->key;
	if(u->r!=NIL) zhong(u->r);
	return ;
}
void qian(node *u){
	cout<<" "<<u->key;
	if(u->l!=NIL) qian(u->l);
	if(u->r!=NIL) qian(u->r);
	return;
}
void print(node *u){
	zhong(u);cout<<endl;
	qian(u);cout<<endl;return;
}
node* find(int num){
	node *p=root;
	 while(p!=NIL&&p->key!=num){
	 	if(num<p->key) p=p->l;
	 	else if(num>p->key) p=p->r;
	 }
	 return p;
}
int main(){
	int cnt;int num;
	scanf("%d",&cnt);
	char order[10];
	vector<int> ::iterator it;
	for(int i=0;i<cnt;i++){
		scanf("%s",&order);
		if(order[0]=='i') {
		scanf("%d",&num);insert(num);
		}
		else if(order[0]=='p') {print(root);
		}else{scanf("%d",&num);
		if(find(num)!=NIL){cout<<"yes"<<endl;
		}else{
			cout<<"no"<<endl;  
		}
			
		}
	}
}

```