#include <iostream>
#include <vector>
#include <string>

#define NIL -1

using namespace std;

class NODE {
public:
	int key, p = NIL, right = NIL, left = NIL;
	NODE(int key) {
		this->key = key;
	}
};

vector<NODE> T;
int root = NIL;

void insert(int z) {
	int y = NIL, x = root;
	while (x != NIL) {
		y = x;
		if (T[z].key < T[x].key) x = T[x].left;
		else x = T[x].right;
	}
	T[z].p = y;
	if (y == NIL) { root = z; }
	else if (T[z].key < T[y].key) { T[y].left = z; }
	else T[y].right = z;
}

void find(int k) {
	int x = root;
	bool t = false;
	while (x != NIL) {
		if (k == T[x].key) { t = true; break; }
		else if (k < T[x].key) x = T[x].left;
		else x = T[x].right;
	}
	if (t == true) cout << "yes" << endl;
	else cout << "no" << endl;
}

void In(int a) {
	if (a != NIL) {
		In(T[a].left);
		cout << " " << T[a].key;
		In(T[a].right);
	}
}

void Pre(int a) {
	if (a != NIL) {
		cout << " " << T[a].key;
		Pre(T[a].left);
		Pre(T[a].right);
	}
}


int main() {
	string cmd;
	int m, cnt = 0, key;
	cin >> m;
	for (int i = 0; i < m; i++) {
		cin >> cmd;
		if (cmd == "insert") {
			cin >> key;
			NODE z(key);
			T.push_back(z);
			insert(cnt);
			cnt++;
		}

		else if (cmd == "find") {
			cin >> key;
			find(key);
		}

		else if (cmd == "print") {
			In(root); cout << endl;
			Pre(root); cout << endl;
		}
	}

	return 0;
}

### Prompt

Create a solution in CPP for 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 <vector>
#include <string>

#define NIL -1

using namespace std;

class NODE {
public:
	int key, p = NIL, right = NIL, left = NIL;
	NODE(int key) {
		this->key = key;
	}
};

vector<NODE> T;
int root = NIL;

void insert(int z) {
	int y = NIL, x = root;
	while (x != NIL) {
		y = x;
		if (T[z].key < T[x].key) x = T[x].left;
		else x = T[x].right;
	}
	T[z].p = y;
	if (y == NIL) { root = z; }
	else if (T[z].key < T[y].key) { T[y].left = z; }
	else T[y].right = z;
}

void find(int k) {
	int x = root;
	bool t = false;
	while (x != NIL) {
		if (k == T[x].key) { t = true; break; }
		else if (k < T[x].key) x = T[x].left;
		else x = T[x].right;
	}
	if (t == true) cout << "yes" << endl;
	else cout << "no" << endl;
}

void In(int a) {
	if (a != NIL) {
		In(T[a].left);
		cout << " " << T[a].key;
		In(T[a].right);
	}
}

void Pre(int a) {
	if (a != NIL) {
		cout << " " << T[a].key;
		Pre(T[a].left);
		Pre(T[a].right);
	}
}


int main() {
	string cmd;
	int m, cnt = 0, key;
	cin >> m;
	for (int i = 0; i < m; i++) {
		cin >> cmd;
		if (cmd == "insert") {
			cin >> key;
			NODE z(key);
			T.push_back(z);
			insert(cnt);
			cnt++;
		}

		else if (cmd == "find") {
			cin >> key;
			find(key);
		}

		else if (cmd == "print") {
			In(root); cout << endl;
			Pre(root); cout << endl;
		}
	}

	return 0;
}

```