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.Scanner;

public class Main {
	public static void main(String[] args) {
		int i, j;
		try (Scanner sc = new Scanner(System.in)) {
			int n = sc.nextInt();
			int q = sc.nextInt();
			ArrayList<Deque<Integer>> array = new ArrayList<>();
			for(i = 0; i < n; i++) {
				array.add(new ArrayDeque<Integer>());
			}
			for(i = 0; i < q; i++) {
				int order_number = sc.nextInt();
				int t = sc.nextInt();
				if(array.get(t).isEmpty() && order_number != 0) {
					continue;
				}
				switch(order_number) {
				case 0 :
					array.get(t).add(sc.nextInt());
					break;
				case 1 :
					System.out.println(array.get(t).peekLast());
					
					break;
				case 2 :
					array.get(t).removeLast();
					break;
				}
//				for (Deque<Integer> output : array) {
//					System.out.print(output);
//				}
			}

		}
	}
}