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


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

Node *root, *NIL;

void insert(int k) {
	Node *y = NIL;
	Node *x = root;
	Node *z;


	z = (Node *)malloc(sizeof(Node));

	z->key = k;
	z->left = NIL;
	z->right = NIL;

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

		}
		else {
			y->right = z;
		}
	}
}

Node* find(Node *x, int k) {

	while (x != NIL && k != x->key) {
		if (k < x->key) {
			x = x->left;
		}
		else {
			x = x->right;
		}
	}
	return x;
}

void inorder(Node *u) {
	if (u == NIL)return;
	inorder(u->left);
	printf(" %d", u->key);
	inorder(u->right);
}

void preorder(Node *u) {
	if (u == NIL)return;
	printf(" %d", u->key);
	preorder(u->left);
	preorder(u->right);

}

int main() {
	int n, i, x;
	string com;

	cin >> n;
	for (i = 0; i < n; i++) {
		cin >> com;
		if (com == "insert") {
			scanf("%d", &x);
			insert(x);
		}
		else if (com == "print") {
			inorder(root);
			printf("\n");
			preorder(root);
			printf("\n");

		}
		else if (com == "find") {
			scanf("%d", &x);
			Node* no=find(root, x);
			if (no == NIL) {
				printf("no\n");
			}
			else {
				printf("yes\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<iostream>
#include<cstdio>
#include<string>
#include<cstdlib>
using namespace std;


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

Node *root, *NIL;

void insert(int k) {
	Node *y = NIL;
	Node *x = root;
	Node *z;


	z = (Node *)malloc(sizeof(Node));

	z->key = k;
	z->left = NIL;
	z->right = NIL;

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

		}
		else {
			y->right = z;
		}
	}
}

Node* find(Node *x, int k) {

	while (x != NIL && k != x->key) {
		if (k < x->key) {
			x = x->left;
		}
		else {
			x = x->right;
		}
	}
	return x;
}

void inorder(Node *u) {
	if (u == NIL)return;
	inorder(u->left);
	printf(" %d", u->key);
	inorder(u->right);
}

void preorder(Node *u) {
	if (u == NIL)return;
	printf(" %d", u->key);
	preorder(u->left);
	preorder(u->right);

}

int main() {
	int n, i, x;
	string com;

	cin >> n;
	for (i = 0; i < n; i++) {
		cin >> com;
		if (com == "insert") {
			scanf("%d", &x);
			insert(x);
		}
		else if (com == "print") {
			inorder(root);
			printf("\n");
			preorder(root);
			printf("\n");

		}
		else if (com == "find") {
			scanf("%d", &x);
			Node* no=find(root, x);
			if (no == NIL) {
				printf("no\n");
			}
			else {
				printf("yes\n");
			}

		}
	}

	return 0;

}

```