#include <stdio.h>
#include <vector>
#include <stack>

using namespace std;

int main()
{
  int n,q,c,t,x;
  vector<stack<int> > a;

  scanf("%d", &n);
  scanf("%d", &q);


  for (int i = 0; i < n; i++) {
    a.emplace_back();
  }


  for (int i = 0; i < q; i++) {
    scanf("%d", &c);
    scanf("%d", &t);
    if (c == 0){
      scanf("%d", &x);
      a[t].push(x);
    }else if (c == 1){
      if (a[t].empty()==0){ //空：１　
        printf("%d\n",a[t].top());
      }
    }else if (c == 2){
      if (a[t].empty()==0){ //空：１　
        a[t].pop();
      }
    }

  }
  return 0;
}


### 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 <stdio.h>
#include <vector>
#include <stack>

using namespace std;

int main()
{
  int n,q,c,t,x;
  vector<stack<int> > a;

  scanf("%d", &n);
  scanf("%d", &q);


  for (int i = 0; i < n; i++) {
    a.emplace_back();
  }


  for (int i = 0; i < q; i++) {
    scanf("%d", &c);
    scanf("%d", &t);
    if (c == 0){
      scanf("%d", &x);
      a[t].push(x);
    }else if (c == 1){
      if (a[t].empty()==0){ //空：１　
        printf("%d\n",a[t].top());
      }
    }else if (c == 2){
      if (a[t].empty()==0){ //空：１　
        a[t].pop();
      }
    }

  }
  return 0;
}


```