#include <algorithm>
#include<iostream>
#include<vector>
#include<deque>
#include<stack>
#include<list>
#include<map>
#include<set>
#include<string>
#include <sstream>

#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<limits.h>

using namespace std;


int main(){
  long ii,jj,kk;
  vector<int> ret;
  stack<int> s[1001];
  
  int n,q,c,t,x;

  cin >> n >> q;

  for(ii=0;ii<q;ii++){
    //cout << ii << endl;
    cin >> c;
    switch(c){

    case 0:
      cin >> t >> x;
      s[t].push(x);
      break;

    case 1:
      cin >> t;
      if(!s[t].empty()){
        cout << s[t].top() << endl;
      }
      break;

    case 2:
      cin >> t;
      if(!s[t].empty()){
        s[t].pop();
      }
      break;

    }
  }
  
  return 0;
}


### Prompt

Generate a Cpp solution 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 <algorithm>
#include<iostream>
#include<vector>
#include<deque>
#include<stack>
#include<list>
#include<map>
#include<set>
#include<string>
#include <sstream>

#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<limits.h>

using namespace std;


int main(){
  long ii,jj,kk;
  vector<int> ret;
  stack<int> s[1001];
  
  int n,q,c,t,x;

  cin >> n >> q;

  for(ii=0;ii<q;ii++){
    //cout << ii << endl;
    cin >> c;
    switch(c){

    case 0:
      cin >> t >> x;
      s[t].push(x);
      break;

    case 1:
      cin >> t;
      if(!s[t].empty()){
        cout << s[t].top() << endl;
      }
      break;

    case 2:
      cin >> t;
      if(!s[t].empty()){
        s[t].pop();
      }
      break;

    }
  }
  
  return 0;
}


```