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;

class Main {

	public static void main(String[] args) {
		ArrayList <Deque<Integer>> deques = new ArrayList <Deque<Integer>>();  //DequeのArrayListを生成
		try (Scanner sc = new Scanner (System.in)){
			int n = sc.nextInt();
			for (int i=0; i<n; i++) {
				deques.add(new ArrayDeque<Integer>());  //n個のdequeを追加
			}
			int times = sc.nextInt();
			for (int i=0; i<times; i++) {
				int request = sc.nextInt();  //命令 0 push 1 top 2 pop
				int deq = sc.nextInt();  //何個目のdeque
				switch (request) {
				//push(deq,x)
				case 0:
					int x = sc.nextInt();
					deques.get(deq).addFirst(x);
					break;
				//top(deq)
				case 1:
					if (deques.get(deq).size() != 0) { //空じゃなかったら
						System.out.println(deques.get(deq).peek());
					}
					break;
				//pop(deq)
				case 2:
					if (deques.get(deq).size() != 0) { //空じゃなかったら
						deques.get(deq).pop();
					}
					break;

				}
			}
		}
	}
}