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.*;

class Main{
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        
        int n = sc.nextInt();   //スタック要素数
        int q = sc.nextInt();   //命令数

        ArrayList<Deque<Integer>> stack = new ArrayList<>();
        for(int i = 0;i < n;i++){
            stack.add(new ArrayDeque<>());
        }

        for(int i = 0;i < q;i++){
            int num = sc.nextInt(); //命令番号
            int t = sc.nextInt();   //指定スタック番号
            switch(num){
                case 0: //push
                    int x = sc.nextInt();
                    stack.get(t).add(x);
                    break;
                case 1: //top
                    if(stack.get(t).size() > 0)
                        System.out.println(stack.get(t).getLast());
                    break;
                case 2: //pop
                    if(stack.get(t).size() > 0)
                        stack.get(t).removeLast();
                    break;
            }
        }
    }
}