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>

using namespace std;
#define MAX 1000

class stack{
    int num;
    stack* head = NULL;
    stack* prev = NULL;
    stack* tail = NULL;
    bool isEmpty = true;
public:
    void push(int x){
        if(isEmpty){
            isEmpty = false;
            num = x;
            return;
        }
        if(head == NULL){
            head = new stack(x);
            head->prev = this;
            this->tail = this;
        }
        else{
            stack* tmp = tail->head;
            tmp->prev = tail;
            tmp->head = new stack(x);
            tail =  tmp;
        }
    }
    
    int top(){
        if(head == NULL){
            return num;
        }
        else{
            return tail->head->num;
        }
    }
    
    void pop(){
        if(head == NULL){
            isEmpty = true;
        }
        else{
            stack* tmp = tail;
            delete tmp->head;
            tmp->head = NULL;
            tail = tail->prev;
        }
    }
    
    bool empty(){
        return isEmpty;
    }
    
    stack(){
        
    }
    
    stack(int num){
        this->num = num;
    }
};

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    int n,q,command,t,x;
    stack myStack[MAX];
    cin >> n >> q;
    
    for (int i = 0;i < q;++i) {
        cin >> command >> t;
        switch(command){
                // push
            case 0:
                cin >> x;
                myStack[t].push(x);
                break;
                // top
            case 1:
                if(!myStack[t].empty()){
                    cout << myStack[t].top() << endl;
                }
                break;
                // clear
            case 2:
                if(!myStack[t].empty()){
                    myStack[t].pop();
                }
                break;
        }
    }
}