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.Scanner;
import java.util.Stack;

public class Main {
	public static void main(String args[]) {
		
		Scanner scan = new Scanner(System.in);
		Stack<Stack<Integer>> st = new Stack<>();
		
		int n = scan.nextInt();
		for(int i=0; i<n; i++) {
			st.add(new Stack<Integer>());
		}
		
		int count = scan.nextInt();
		
		for(int i=0; i<count; i++) {
			int query = scan.nextInt();
			int a = scan.nextInt();

			switch(query) {
			
			case 0:
				int b = scan.nextInt();
				st.get(a).push(b);
				break;
				
			case 1:
				if(!st.get(a).isEmpty())
				System.out.println(st.get(a).peek());
				
				break;
				
			case 2:
				if(!st.get(a).isEmpty())
				st.get(a).pop();
				
				break;
			}
		}
	}
}