#include<bits/stdc++.h>

using namespace std;

vector<stack<int> > st;

void f0()
{
    int p1,p2;
    cin>>p1>>p2;

    st[p1].push(p2);
}

void f1()
{
    int p;
    cin>>p;

    if(!st[p].empty())
        cout<<st[p].top()<<endl;
}

void f2()
{
    int p;
    cin>>p;

    if(!st[p].empty()) st[p].pop();
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n,q;

    cin>>n>>q;

    st.resize(n);

    while(q--)
    {
        int op;
        cin>>op;

        switch(op)
        {
            case 0:f0();break;
            case 1:f1();break;
            case 2:f2();break;
        }
    }


    return 0;
}


### Prompt

In cpp, your task is to solve 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<bits/stdc++.h>

using namespace std;

vector<stack<int> > st;

void f0()
{
    int p1,p2;
    cin>>p1>>p2;

    st[p1].push(p2);
}

void f1()
{
    int p;
    cin>>p;

    if(!st[p].empty())
        cout<<st[p].top()<<endl;
}

void f2()
{
    int p;
    cin>>p;

    if(!st[p].empty()) st[p].pop();
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n,q;

    cin>>n>>q;

    st.resize(n);

    while(q--)
    {
        int op;
        cin>>op;

        switch(op)
        {
            case 0:f0();break;
            case 1:f1();break;
            case 2:f2();break;
        }
    }


    return 0;
}


```