#include <iostream>
#include <iomanip>
#include<vector>
#include <algorithm>
#include <queue>
#include<string>
#include <map>
#include <cmath>
#include <deque>
#include <list>
#include <stack>
typedef long long ll;
typedef long double ld;

#define REP(i,s,n)for(ll i=s;i<(n);++i)
#define rep(i,n) REP(i,0,n)
using namespace std;
int S[1000] = { 0 };


int main() {
	ll N,Q;
	cin >>N>>Q;
	stack<ll>Y[N];
	rep(i, Q) {
		int a;
		cin >> a;
		if (a == 0) {
			int b,c;
			cin >> b>>c;
			Y[b].push(c);
		}
		else if (a == 1) {
			int b;
			cin >> b;
			if (0 < Y[b].size()) cout << Y[b].top() << endl;
		}
		else {
			int b;
			cin >> b;
			if (0 < Y[b].size()) Y[b].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
#include <iostream>
#include <iomanip>
#include<vector>
#include <algorithm>
#include <queue>
#include<string>
#include <map>
#include <cmath>
#include <deque>
#include <list>
#include <stack>
typedef long long ll;
typedef long double ld;

#define REP(i,s,n)for(ll i=s;i<(n);++i)
#define rep(i,n) REP(i,0,n)
using namespace std;
int S[1000] = { 0 };


int main() {
	ll N,Q;
	cin >>N>>Q;
	stack<ll>Y[N];
	rep(i, Q) {
		int a;
		cin >> a;
		if (a == 0) {
			int b,c;
			cin >> b>>c;
			Y[b].push(c);
		}
		else if (a == 1) {
			int b;
			cin >> b;
			if (0 < Y[b].size()) cout << Y[b].top() << endl;
		}
		else {
			int b;
			cin >> b;
			if (0 < Y[b].size()) Y[b].pop();
		}
	}
	return 0;
}




```