#include <iostream>
#include <vector>
#include <stack>
using namespace std;

#define sint stack<int>

int main() {
	vector<sint> vec; int an, tn, fn1, fn2, en;
	cin >> an >> tn; vec.resize(an);
	for (int i = 0; i < tn; i++) {
		cin >> fn1;
		switch (fn1) {
		case 0: cin >> fn2 >> en; vec[fn2].push(en); break;
		case 1: cin >> fn2; 
			if (vec[fn2].empty()) continue;
			else cout << vec[fn2].top() << endl;
			break;
		case 2: cin >> fn2;
			if (vec[fn2].empty()) continue;
			else vec[fn2].pop();
			break;
		}
	}
	return 0;
}

### Prompt

Develop a solution in CPP 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 <iostream>
#include <vector>
#include <stack>
using namespace std;

#define sint stack<int>

int main() {
	vector<sint> vec; int an, tn, fn1, fn2, en;
	cin >> an >> tn; vec.resize(an);
	for (int i = 0; i < tn; i++) {
		cin >> fn1;
		switch (fn1) {
		case 0: cin >> fn2 >> en; vec[fn2].push(en); break;
		case 1: cin >> fn2; 
			if (vec[fn2].empty()) continue;
			else cout << vec[fn2].top() << endl;
			break;
		case 2: cin >> fn2;
			if (vec[fn2].empty()) continue;
			else vec[fn2].pop();
			break;
		}
	}
	return 0;
}

```