#include <algorithm>
#include <iostream>
#include <stack>
#include <vector>

using namespace std;

typedef long long ll;
typedef pair<ll, ll> P;

#define MOD 1000000007
#define REP(i, N) for (int i = 0; i < N; ++i)
#define REP1(i, N) for (int i = 1; i <= N; ++i)
#define RREP(i, N) for (int i = N - 1; i >= 0; --i)
#define ALL(a) a.begin(), a.end()

int main() {
  int n, q;
  cin >> n >> q;
  stack<int> st[n];
  REP(i, q) {
    int num, t;
    cin >> num >> t;
    if (num == 0) {
      int x;
      cin >> x;
      st[t].push(x);
    } else if (num == 1) {
      if (st[t].size()) {
        cout << st[t].top() << endl;
      }
    } else {
      if (st[t].size()) {
        st[t].pop();
      }
    }
  }
  return 0;
}

### Prompt

Please formulate a CPP solution to the following 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

### Response

```cpp
#include <algorithm>
#include <iostream>
#include <stack>
#include <vector>

using namespace std;

typedef long long ll;
typedef pair<ll, ll> P;

#define MOD 1000000007
#define REP(i, N) for (int i = 0; i < N; ++i)
#define REP1(i, N) for (int i = 1; i <= N; ++i)
#define RREP(i, N) for (int i = N - 1; i >= 0; --i)
#define ALL(a) a.begin(), a.end()

int main() {
  int n, q;
  cin >> n >> q;
  stack<int> st[n];
  REP(i, q) {
    int num, t;
    cin >> num >> t;
    if (num == 0) {
      int x;
      cin >> x;
      st[t].push(x);
    } else if (num == 1) {
      if (st[t].size()) {
        cout << st[t].top() << endl;
      }
    } else {
      if (st[t].size()) {
        st[t].pop();
      }
    }
  }
  return 0;
}

```