Problem: Stack is a container of elements that are inserted and deleted according to LIFO (Last In First Out).

For $n$ stack $S_i$ ($i = 0, 1, ..., n-1$), perform a sequence of the following operations.

* push($t$, $x$): Insert an integer $x$ to $S_t$.
* top($t$): Report the value which should be deleted next from $S_t$. If $S_t$ is empty, do nothing.
* pop($t$): Delete an element from $S_t$. If $S_t$ is empty, do nothing.



In the initial state, all stacks are empty.

Constraints

* $1 \leq n \leq 1,000$
* $1 \leq q \leq 200,000$
* $-1,000,000,000 \leq x \leq 1,000,000,000$

Input

The input is given in the following format.


$n \; q$
$query_1$
$query_2$
:
$query_q$


Each query $query_i$ is given by


0 $t$ $x$


or


1 $t$


or


2 $t$


where the first digits 0, 1 and 2 represent push, top and pop operations respectively.

Output

For each top operation, print an integer in a line.

Example

Input

3 9
0 0 1
0 0 2
0 0 3
0 2 4
0 2 5
1 0
1 2
2 0
1 0


Output

3
5
2

Solution (in JAVA):

Solution:
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Deque;
import java.util.List;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		// TODO 自動生成されたメソッド・スタブ
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int q = sc.nextInt();

		//Deque型のListを生成
		List<Deque<Integer>> list = new ArrayList<Deque<Integer>>();
		for(int i = 0; i < n; i++) {
			list.add(new ArrayDeque<Integer>());
		}

		for(int i = 0; i < q; i++)	{
			int query = sc.nextInt();
			int t = sc.nextInt();

			switch(query){
			case 0:
				int x = sc.nextInt();
				list.get(t).add(x);  //listの中のt番目の要素を取得し、それにaddする
				break;

			case 1:
				if(!(list.get(t).isEmpty())) {
					System.out.println(list.get(t).peekLast());
				}
				break;

			case 2:
				list.get(t).pollLast();
				break;
			}
		}
	}

}