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 CPP):

Solution:
#include<iostream>
#include<stack>
#include<cstdio>
#include<vector>

using namespace std;

int main(){
    int num_stack,num_query;
    int command,stack_id,val;
    
    scanf("%d %d",&num_stack,&num_query);
    stack<int> stack_array[num_stack];
    for(int i = 0; i < num_query;i++){
        scanf("%d %d",&command,&stack_id);
        if(command == 0){
            scanf("%d",&val);
            stack_array[stack_id].push(val);
        }else if(command == 1){
            if(!stack_array[stack_id].empty()) printf("%d\n",stack_array[stack_id].top());
        }else{
            if(!stack_array[stack_id].empty()) stack_array[stack_id].pop();
        }
    }
    
    
    
    return 0;
}