#include <iostream>
using namespace std;

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

#define MAX 500000

Node* root;

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

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

void print() {
	in(root);
	cout << endl;
	pre(root);
	cout << endl;
}

void insert(int key) {
	Node* y = NULL;
	Node* x = root;
	Node* z = new Node;
	z->key = key;
	z->parent = z->left = z->right = NULL;

	if (root == NULL) {
		root = z;
		return;
	}

	while (x != NULL) {
		y = x;
		if (key < x->key)x = y->left;
		else if (key > x->key)x = y->right;
	}
	if (key < y->key) {
		y->left = z;
		z->parent = y;
	}
	else {
		y->right = z;
		z->parent = y;
	}
}

Node* find(int key) {
	Node* p = root;
	while (p!=NULL) {
		if (key < p->key)p = p->left;
		else if (key > p->key)p = p->right;
		else return p;
	}
	return NULL;
}

int main() {
	int m;
	cin >> m;
	for (int i = 0; i < m; i++) {
		char buf[20];
		cin >> buf;
		if (buf[0] == 'i') {
			int k;
			cin >> k;
			insert(k);
		}
		else if (buf[0] == 'f') {
			int k;
			cin >> k;
			if (find(k) != NULL)cout << "yes" << endl;
			else cout << "no" << endl;
		}
		else {
			print();
		}
	}
	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>
using namespace std;

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

#define MAX 500000

Node* root;

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

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

void print() {
	in(root);
	cout << endl;
	pre(root);
	cout << endl;
}

void insert(int key) {
	Node* y = NULL;
	Node* x = root;
	Node* z = new Node;
	z->key = key;
	z->parent = z->left = z->right = NULL;

	if (root == NULL) {
		root = z;
		return;
	}

	while (x != NULL) {
		y = x;
		if (key < x->key)x = y->left;
		else if (key > x->key)x = y->right;
	}
	if (key < y->key) {
		y->left = z;
		z->parent = y;
	}
	else {
		y->right = z;
		z->parent = y;
	}
}

Node* find(int key) {
	Node* p = root;
	while (p!=NULL) {
		if (key < p->key)p = p->left;
		else if (key > p->key)p = p->right;
		else return p;
	}
	return NULL;
}

int main() {
	int m;
	cin >> m;
	for (int i = 0; i < m; i++) {
		char buf[20];
		cin >> buf;
		if (buf[0] == 'i') {
			int k;
			cin >> k;
			insert(k);
		}
		else if (buf[0] == 'f') {
			int k;
			cin >> k;
			if (find(k) != NULL)cout << "yes" << endl;
			else cout << "no" << endl;
		}
		else {
			print();
		}
	}
	return 0;
}

```