#include <bits/stdc++.h>
#include <vector>
 
typedef long long ll;
#define SIZE_OF_ARRAY(array) (sizeof(array)/sizeof(array[0]))
#define rep(i, n) for (int i = 0; i < (n); ++i)
const double PI=3.14159265358979323846;
 
using namespace std;
 
int main() {

	int n, q;
	cin >> n >> q;

	stack<int> s[n];

	int a, b, c;

	rep(i, q) {
		cin >> a >> b;

		switch(a) {

			case 0:
				cin >> c;
				s[b].push(c);
				break;

			case 1:
				if (!s[b].empty())
					cout << s[b].top() << endl;
				break;
			
			case 2:
				if (!s[b].empty())
					s[b].pop();
				break;
		}
	}


}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 <vector>
 
typedef long long ll;
#define SIZE_OF_ARRAY(array) (sizeof(array)/sizeof(array[0]))
#define rep(i, n) for (int i = 0; i < (n); ++i)
const double PI=3.14159265358979323846;
 
using namespace std;
 
int main() {

	int n, q;
	cin >> n >> q;

	stack<int> s[n];

	int a, b, c;

	rep(i, q) {
		cin >> a >> b;

		switch(a) {

			case 0:
				cin >> c;
				s[b].push(c);
				break;

			case 1:
				if (!s[b].empty())
					cout << s[b].top() << endl;
				break;
			
			case 2:
				if (!s[b].empty())
					s[b].pop();
				break;
		}
	}


}

```