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 <stdio.h>
#include <stack>
using namespace std;

int main(){
    int n,q,query,x,t;;
    scanf("%d %d", &n,&q);
    stack<int> A[n];
    for (int i = 0; i < q; i++)
    {        
        scanf("%d %d", &query,&t);
        switch (query){
            case 0:
                scanf("%d",&x);
                A[t].push(x);                       
                break;
            case 1:
                if (!A[t].empty()) printf("%d\n",A[t].top());
                break;
            case 2:
                if (!A[t].empty()) A[t].pop();
                break;
            default:
                break;
        }
    }    
    return 0;
}