#include <iostream>
#include <vector>

using namespace std;

int main()
{
    long long n,q;
    cin>>n>>q;
    vector < vector<int> > a(n);
    while(q--)
    {
        int b;
        cin>>b;
        if(b==0)
        {
           long long t,x;
           cin>>t>>x;
           a[t].push_back(x);
        }
        else if(b==1)
        {
            long long t;
            cin>>t;
            if(!a[t].empty())
            {
                long long i=a[t].size()-1;
                cout<<a[t][i]<<endl;
            }
        }
        else
        {
            long long t;
            cin>>t;
            if(!a[t].empty())
            {
                a[t].pop_back();
            }
        }
    }
    return 0;
}


### Prompt

Your challenge is to write 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 <iostream>
#include <vector>

using namespace std;

int main()
{
    long long n,q;
    cin>>n>>q;
    vector < vector<int> > a(n);
    while(q--)
    {
        int b;
        cin>>b;
        if(b==0)
        {
           long long t,x;
           cin>>t>>x;
           a[t].push_back(x);
        }
        else if(b==1)
        {
            long long t;
            cin>>t;
            if(!a[t].empty())
            {
                long long i=a[t].size()-1;
                cout<<a[t][i]<<endl;
            }
        }
        else
        {
            long long t;
            cin>>t;
            if(!a[t].empty())
            {
                a[t].pop_back();
            }
        }
    }
    return 0;
}


```