#include<iostream>
#include<algorithm>
#include<stack>
#include<vector>
#include<fstream>
using namespace std;
    
#define NUM 10000
stack<int> s[NUM];

int main(){

    int n, q, order, t;
    cin >> n >> q;
    for(int i=0;i<q;i++){
        cin >> order >> t;
        switch (order)
        {
        case 0:
            int num;
            cin >> num;
            s[t].push(num);
            break;
        case 1:
            if(!(s[t].empty())) cout << s[t].top() << endl;
            break;
        case 2:
            if(!(s[t].empty()))s[t].pop();
            break;
        default:
            break;
        }
    }
    return 0;
}

### Prompt

Create a solution in Cpp for the following 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

### Response

```cpp
#include<iostream>
#include<algorithm>
#include<stack>
#include<vector>
#include<fstream>
using namespace std;
    
#define NUM 10000
stack<int> s[NUM];

int main(){

    int n, q, order, t;
    cin >> n >> q;
    for(int i=0;i<q;i++){
        cin >> order >> t;
        switch (order)
        {
        case 0:
            int num;
            cin >> num;
            s[t].push(num);
            break;
        case 1:
            if(!(s[t].empty())) cout << s[t].top() << endl;
            break;
        case 2:
            if(!(s[t].empty()))s[t].pop();
            break;
        default:
            break;
        }
    }
    return 0;
}

```