Problem: 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

Solution (in CPP):

Solution:
#define _USE_MATH_DEFINES
#define _CRT_SECURE_NO_WARNINGS

#include <iostream>
#include <string>
#include <sstream>
#include <iomanip>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <stack>
#include <queue>
#include <vector>
#include <list>
#include <set>
#include <map>
#include <bitset>
#include <utility>
#include <numeric>

using namespace std;

using ll = long long;
using ull = unsigned long long;

const ll inf = (ll)1e9;
const ll mod = (ll)1e9 + 7;

#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(), v.rend()

#define print(s) cout << s;
#define println(s) cout << s << endl;
#define printd(s, f) cout << fixed << setprecision(f) << s << endl;


int main()
{
	int n, m;
	cin >> n >> m;

	vector<stack<int>> s(n);

	for (int i = 0; i < m; i++)
	{
		int q, t;
		cin >> q >> t;

		int x;
		switch (q)
		{
			case 0:
				cin >> x;
				s[t].push(x);
				break;

			case 1:
				if (!s[t].empty())
				{
					println(s[t].top());
				}
				break;

			case 2:
				if (!s[t].empty())
				{
					s[t].pop();
				}
				break;
		}
	}
}