#include <cstdio>
#include <cstdlib>
#include <string>
#include <iostream>
using namespace std;

struct BINNODE {
	int key;
	BINNODE *parent;
	BINNODE *left;
	BINNODE *right;

	BINNODE (int k) {
		key=k;
		parent=nullptr;
		left=nullptr;
		right=nullptr;
	}
};

void preorderwalk(BINNODE* root) {
	if(root==nullptr) return;
	printf(" %d",root->key);
	preorderwalk(root->left);
	preorderwalk(root->right);
	return;
}

void inorderwalk(BINNODE* root) {
	if(root==nullptr) return;
	inorderwalk(root->left);
	printf(" %d",root->key);
	inorderwalk(root->right);
	return;
}

bool findnode(int v, BINNODE* root) {
	if(root == nullptr) return false;
	if(root->key ==v ) return true;
	return (v< root->key)?findnode(v,root->left) : findnode(v,root->right);
}

int main() {
	int m,v;
	int i;
	string cmd;
	BINNODE* root=nullptr;
	BINNODE* x;
	BINNODE* y;
	BINNODE* z;
	cin >> m;
	for(i=0; i<m; ++i) {
		cin >> cmd;
		if(cmd=="find") {
			cin >> v;
			printf(findnode(v,root) ? "yes\n" : "no\n");
		} else if(cmd=="insert") {
			cin >> v;
			z=new BINNODE(v);
			y=nullptr;
			x=root;
			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;
			}
		} else {
			inorderwalk(root);
			printf("\n");
			preorderwalk(root);
			printf("\n");
		}
	}
	return 0;
}
### 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 <cstdio>
#include <cstdlib>
#include <string>
#include <iostream>
using namespace std;

struct BINNODE {
	int key;
	BINNODE *parent;
	BINNODE *left;
	BINNODE *right;

	BINNODE (int k) {
		key=k;
		parent=nullptr;
		left=nullptr;
		right=nullptr;
	}
};

void preorderwalk(BINNODE* root) {
	if(root==nullptr) return;
	printf(" %d",root->key);
	preorderwalk(root->left);
	preorderwalk(root->right);
	return;
}

void inorderwalk(BINNODE* root) {
	if(root==nullptr) return;
	inorderwalk(root->left);
	printf(" %d",root->key);
	inorderwalk(root->right);
	return;
}

bool findnode(int v, BINNODE* root) {
	if(root == nullptr) return false;
	if(root->key ==v ) return true;
	return (v< root->key)?findnode(v,root->left) : findnode(v,root->right);
}

int main() {
	int m,v;
	int i;
	string cmd;
	BINNODE* root=nullptr;
	BINNODE* x;
	BINNODE* y;
	BINNODE* z;
	cin >> m;
	for(i=0; i<m; ++i) {
		cin >> cmd;
		if(cmd=="find") {
			cin >> v;
			printf(findnode(v,root) ? "yes\n" : "no\n");
		} else if(cmd=="insert") {
			cin >> v;
			z=new BINNODE(v);
			y=nullptr;
			x=root;
			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;
			}
		} else {
			inorderwalk(root);
			printf("\n");
			preorderwalk(root);
			printf("\n");
		}
	}
	return 0;
}
```