#include <bits/stdc++.h>
#include <deque>
using namespace std;
#define rep(i,n) for (int i = 0; i < (n); i++)
#define repd(i,a,b) for (int i = (a); i < (b); i++)
typedef long long ll;
int main(void)
{
	int n, q, query, t, x;
	cin >> n >> q;
	vector<deque<int> > S(n);
	
	rep(i, q)
	{
		cin >> query >> t;
		if (query == 0)
		{
			cin >> x;
			S[t].push_back(x);
		}
		else if (query == 1)
		{
		    if (S[t].size())
		    {
			    cout << S[t].back() << endl;
		    }
		}
		else
		{
		    if (S[t].size())
			{
			    S[t].pop_back();
			}
		}
	}
}

### Prompt

Create a solution in cpp for 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>
#include <deque>
using namespace std;
#define rep(i,n) for (int i = 0; i < (n); i++)
#define repd(i,a,b) for (int i = (a); i < (b); i++)
typedef long long ll;
int main(void)
{
	int n, q, query, t, x;
	cin >> n >> q;
	vector<deque<int> > S(n);
	
	rep(i, q)
	{
		cin >> query >> t;
		if (query == 0)
		{
			cin >> x;
			S[t].push_back(x);
		}
		else if (query == 1)
		{
		    if (S[t].size())
		    {
			    cout << S[t].back() << endl;
		    }
		}
		else
		{
		    if (S[t].size())
			{
			    S[t].pop_back();
			}
		}
	}
}

```