#include <iostream>
#include <stack>
#include <vector>

int main() {
  int n, q;
  std::cin >> n >> q;
  std::vector<std::stack<int>> s(n);

  for (int i = 0; i < q; ++i) {
    int c, t;
    std::cin >> c >> t;
    switch (c) {
    case 0:
      int x;
      std::cin >> x;
      s.at(t).push(x);
      break;
    case 1:
      if (s.at(t).size() > 0)
        std::cout << s.at(t).top() << std::endl;
      break;
    case 2:
      if (s.at(t).size() > 0)
        s.at(t).pop();
      break;
    }
  }
}


### Prompt

Please create a solution in cpp to 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 <stack>
#include <vector>

int main() {
  int n, q;
  std::cin >> n >> q;
  std::vector<std::stack<int>> s(n);

  for (int i = 0; i < q; ++i) {
    int c, t;
    std::cin >> c >> t;
    switch (c) {
    case 0:
      int x;
      std::cin >> x;
      s.at(t).push(x);
      break;
    case 1:
      if (s.at(t).size() > 0)
        std::cout << s.at(t).top() << std::endl;
      break;
    case 2:
      if (s.at(t).size() > 0)
        s.at(t).pop();
      break;
    }
  }
}


```