#include<iostream>
#include<string>
using namespace std;
struct Node {
	int key;
	Node* left, * right, * parent;
};
Node* root, * NILL;
void insert(int k) {
	Node* p = NILL;
	Node* id = root;
	Node* z;
	z = new Node;
	z->key = k;
	z->left = NILL;
	z->right = NILL;
	while (id != NILL) {
		p = id;
		if (z->key < id->key) {
			id = id->left;
		}
		else {
			id = id->right;
		}
	}
	z->parent = p;
	if (p == NILL) {
		root = z;
	}
	else if (z->key < p->key) {
		p->left = z;
	}
	else if (z->key > p->key) {
		p->right = z;
	}
}
void find(int k) {
	Node* x = root;
	while (x != NILL && k != x->key) {
		if (k < x->key) {
			x = x->left;
		}
		else {
			x = x->right;
		}
	}
	if (x != NILL) {
		cout<<"yes"<<endl;;
	}
	else {
		cout << "no" << endl;
	}
}
void Preorder(Node* u) {
	if (u == NILL) return;
	else {
		cout << " " << u->key;
		Preorder(u->left);
		Preorder(u->right);
	}
}
void Inorder(Node* u) {
	if (u == NILL) return;
	else {
		Inorder(u->left);
		cout << " " << u->key;
		Inorder(u->right);
	}

}
int main() {
	int n, value;
	string command;
	cin >> n;
	for (int i = 0; i < n; i++) {
		cin >> command;
		if (command == "insert") {
			cin >> value;
			insert(value);
		}
		else if (command == "print") {
			Inorder(root);
			cout << endl;
			Preorder(root);
			cout << endl;
		}
		else if (command == "find") {
			cin >> value;
			find(value);
		}
	}
	return 0;
}

### Prompt

Your challenge is to write 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<string>
using namespace std;
struct Node {
	int key;
	Node* left, * right, * parent;
};
Node* root, * NILL;
void insert(int k) {
	Node* p = NILL;
	Node* id = root;
	Node* z;
	z = new Node;
	z->key = k;
	z->left = NILL;
	z->right = NILL;
	while (id != NILL) {
		p = id;
		if (z->key < id->key) {
			id = id->left;
		}
		else {
			id = id->right;
		}
	}
	z->parent = p;
	if (p == NILL) {
		root = z;
	}
	else if (z->key < p->key) {
		p->left = z;
	}
	else if (z->key > p->key) {
		p->right = z;
	}
}
void find(int k) {
	Node* x = root;
	while (x != NILL && k != x->key) {
		if (k < x->key) {
			x = x->left;
		}
		else {
			x = x->right;
		}
	}
	if (x != NILL) {
		cout<<"yes"<<endl;;
	}
	else {
		cout << "no" << endl;
	}
}
void Preorder(Node* u) {
	if (u == NILL) return;
	else {
		cout << " " << u->key;
		Preorder(u->left);
		Preorder(u->right);
	}
}
void Inorder(Node* u) {
	if (u == NILL) return;
	else {
		Inorder(u->left);
		cout << " " << u->key;
		Inorder(u->right);
	}

}
int main() {
	int n, value;
	string command;
	cin >> n;
	for (int i = 0; i < n; i++) {
		cin >> command;
		if (command == "insert") {
			cin >> value;
			insert(value);
		}
		else if (command == "print") {
			Inorder(root);
			cout << endl;
			Preorder(root);
			cout << endl;
		}
		else if (command == "find") {
			cin >> value;
			find(value);
		}
	}
	return 0;
}

```