#include "bits/stdc++.h"
using namespace std;
using ll = long long;
using P = pair<int, int>;
#define PI acos(-1)
#define Mod (int)1000000007
#define INFTY (int)INT_MAX
#define Rep(i, n) for (int i = 0; i < (int)(n); i++)
#define BitRep(i, n) for (int i = 0; i < (int)(1 << n); i++)
#define All(vec) vec.begin(), vec.end()
#define Sum(vec) accumulate(vec.begin(), vec.end(), 0)
#define Sort(vec) sort(vec.begin(), vec.end())
#define Reverse(vec) reverse(vec.begin(), vec.end())
#define Count(vec, x) count(vec.begin(), vec.end(), x)
#define Next_permutation(vec) next_permutation(vec.begin(), vec.end())
#define Find(S, s) S.find(s) != string::npos

int main() {
	int N, q;
	cin >> N >> q;
	vector<stack<int>> S(N);
	Rep(i, q) {
		int n;
		cin >> n;
		if (n == 0) {
			int t, x;
			cin >> t >> x;
			S[t].push(x);
		}
		else if (n == 1) {
			int t;
			cin >> t;
			if (!S[t].empty()) {
				cout << S[t].top() << endl;
			}
		}
		else if (n == 2) {
			int t;
			cin >> t;
			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
#include "bits/stdc++.h"
using namespace std;
using ll = long long;
using P = pair<int, int>;
#define PI acos(-1)
#define Mod (int)1000000007
#define INFTY (int)INT_MAX
#define Rep(i, n) for (int i = 0; i < (int)(n); i++)
#define BitRep(i, n) for (int i = 0; i < (int)(1 << n); i++)
#define All(vec) vec.begin(), vec.end()
#define Sum(vec) accumulate(vec.begin(), vec.end(), 0)
#define Sort(vec) sort(vec.begin(), vec.end())
#define Reverse(vec) reverse(vec.begin(), vec.end())
#define Count(vec, x) count(vec.begin(), vec.end(), x)
#define Next_permutation(vec) next_permutation(vec.begin(), vec.end())
#define Find(S, s) S.find(s) != string::npos

int main() {
	int N, q;
	cin >> N >> q;
	vector<stack<int>> S(N);
	Rep(i, q) {
		int n;
		cin >> n;
		if (n == 0) {
			int t, x;
			cin >> t >> x;
			S[t].push(x);
		}
		else if (n == 1) {
			int t;
			cin >> t;
			if (!S[t].empty()) {
				cout << S[t].top() << endl;
			}
		}
		else if (n == 2) {
			int t;
			cin >> t;
			if (!S[t].empty()) {
				S[t].pop();
			}
		}
	}
	return 0;
}

```