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);
		
		int n=Integer.parseInt(in.next()),q=Integer.parseInt(in.next());
		mystack A[]=new mystack[n];
		for(int i=0;i<n;i++)A[i]=new mystack();
		
		for(int i=0;i<q;i++){
			int query=Integer.parseInt(in.next());
			
			switch(query){
			case 0:
				int t1=Integer.parseInt(in.next()),x=Integer.parseInt(in.next());
				A[t1].stk.push(x);
				break;
				
			case 1:
				int t2=Integer.parseInt(in.next());
				if(!A[t2].stk.isEmpty())out.println(A[t2].stk.peek());
				break;
				
			case 2:
				int t3=Integer.parseInt(in.next());
				if(!A[t3].stk.isEmpty())A[t3].stk.pop();
				break;
			}
		}
		out.flush();
	}

}

class mystack{
	Stack<Integer> stk;
	
	mystack(){
		stk=new Stack<>();
	}
}