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.*;
import java.io.*;

public class Main {

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);

		ArrayList<Stack<Integer>> stack = new ArrayList<Stack<Integer>>();

		int n = in.nextInt(), loop = in.nextInt();
		for(int i = 0; i < n ; i++) {
			Stack<Integer> stk = new Stack<>();
			stack.add(stk);
		}

		for(int i = 0; i < loop; i++) {
			int odr = Integer.parseInt(in.next()), index = Integer.parseInt(in.next());

			switch(odr) {
				case 0:
					stack.get(index).push(Integer.parseInt(in.next()));
					break;

				case 1:
					if(!stack.get(index).isEmpty()) {
						out.println(stack.get(index).peek());
					}
					break;

				case 2:
					if(!stack.get(index).isEmpty()) {
						stack.get(index).pop();
					}
					break;
			}
		}
		out.flush();
		in.close();

	}

}