#include<iostream>
#include<algorithm>
#include<vector>
#include<numeric>
#include<string>
#include<cmath>
#include<set>
#include<queue>
#include<deque>
#include<bitset>
#include<iomanip>
#include<cctype>
#include<map>

#define rep(i,n) for(int (i)=0;(i)<(n);(i)++)
#define rep2(i,a,b) for (int (i)=a;(i)<(b);(i)++)
#define all(x) (x).begin(),(x).end()

using namespace std;
const int inf = 1001001000;
const long long int Inf = 1001001001001001000;

void print(vector<vector<int>> a){
	for (int i = 0; i < a.size(); i++)
	{
		for (int j=0;j<a[i].size();j++){
			cout << a[i][j] << " ";
		}
		cout << endl;
	}
}

void print(vector<vector<long long int>> a){
	for (int i=0;i<a.size();i++){
		for (int j=0;j<a[i].size();j++){
			cout << a[i][j] << " ";
		}
		cout << endl;
	}
}

void print(vector<int> a){
    int n = a.size();
    for (int j=0;j<n;j++) {
        if (j != n-1) cout << a[j] << " ";
        else cout << a[j] << endl;
    }
}

void print(set<int> a){
	for (auto x:a)
		cout << x << " ";
	cout << endl;
}



int main()
{
	int n, q;
	cin >> n >> q;
	vector<vector<int>> a(n);
	rep(_,q){
		int c;
		cin >> c;
		if(c == 0){
			int t, x;
			cin >> t >> x;
			a[t].push_back(x);
		} else if (c == 2){
			int t;
			cin >> t;
			if (a[t].size())
				a[t].pop_back();
		} else {
			int t;
			cin >> t;
			if (a[t].size())
				cout << a[t].back() << endl;
		}
	}

	return 0;
}

### 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<iostream>
#include<algorithm>
#include<vector>
#include<numeric>
#include<string>
#include<cmath>
#include<set>
#include<queue>
#include<deque>
#include<bitset>
#include<iomanip>
#include<cctype>
#include<map>

#define rep(i,n) for(int (i)=0;(i)<(n);(i)++)
#define rep2(i,a,b) for (int (i)=a;(i)<(b);(i)++)
#define all(x) (x).begin(),(x).end()

using namespace std;
const int inf = 1001001000;
const long long int Inf = 1001001001001001000;

void print(vector<vector<int>> a){
	for (int i = 0; i < a.size(); i++)
	{
		for (int j=0;j<a[i].size();j++){
			cout << a[i][j] << " ";
		}
		cout << endl;
	}
}

void print(vector<vector<long long int>> a){
	for (int i=0;i<a.size();i++){
		for (int j=0;j<a[i].size();j++){
			cout << a[i][j] << " ";
		}
		cout << endl;
	}
}

void print(vector<int> a){
    int n = a.size();
    for (int j=0;j<n;j++) {
        if (j != n-1) cout << a[j] << " ";
        else cout << a[j] << endl;
    }
}

void print(set<int> a){
	for (auto x:a)
		cout << x << " ";
	cout << endl;
}



int main()
{
	int n, q;
	cin >> n >> q;
	vector<vector<int>> a(n);
	rep(_,q){
		int c;
		cin >> c;
		if(c == 0){
			int t, x;
			cin >> t >> x;
			a[t].push_back(x);
		} else if (c == 2){
			int t;
			cin >> t;
			if (a[t].size())
				a[t].pop_back();
		} else {
			int t;
			cin >> t;
			if (a[t].size())
				cout << a[t].back() << endl;
		}
	}

	return 0;
}

```