// 失敗するからこそ そこから立ち向かって行く強さがあってそんな強さが本当の強さだと私は思うから
// ぜったいあきらめない
#include<bits/stdc++.h>
using namespace std;

#if defined(DAIJOBU)
#include "/home/ichigo/debug.hpp"
#define deb(x...) cerr << "[" << #x << "] = ["; _print(x)
#else
#define deb(x...)
#endif
#define FIO {ios::sync_with_stdio(0);cin.tie(nullptr);cout.tie(nullptr);}



int main(){
	FIO
	
	stack<int> s[1100];
	int n, q;
	cin >> n >> q;
	while(q--){
		int op, t;
		cin >> op >> t;
		if(op == 0){
			int x;
			cin >> x;
			s[t].push(x);
		}else if(op == 1){
			if(s[t].empty());
			else cout << s[t].top() << '\n';
		}else{
			if(!s[t].empty())
				s[t].pop();
		}
	}
	return 0;
}


### Prompt

Construct a cpp code solution to the problem outlined:

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
// 失敗するからこそ そこから立ち向かって行く強さがあってそんな強さが本当の強さだと私は思うから
// ぜったいあきらめない
#include<bits/stdc++.h>
using namespace std;

#if defined(DAIJOBU)
#include "/home/ichigo/debug.hpp"
#define deb(x...) cerr << "[" << #x << "] = ["; _print(x)
#else
#define deb(x...)
#endif
#define FIO {ios::sync_with_stdio(0);cin.tie(nullptr);cout.tie(nullptr);}



int main(){
	FIO
	
	stack<int> s[1100];
	int n, q;
	cin >> n >> q;
	while(q--){
		int op, t;
		cin >> op >> t;
		if(op == 0){
			int x;
			cin >> x;
			s[t].push(x);
		}else if(op == 1){
			if(s[t].empty());
			else cout << s[t].top() << '\n';
		}else{
			if(!s[t].empty())
				s[t].pop();
		}
	}
	return 0;
}


```