#include <bits/stdc++.h>
using namespace std;
using ll = long long;

#define rep(i,s,n) for(ll i = (s); i < (n); i++)
#define rep0(i,n) rep(i,0,n)
#define rep1(i,n) rep(i,1,n+1)

#define BR "\n"

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

  stack<int> S[n];

  rep0(i, q) {
    int query; cin >> query;
    if (query == 0) {
      int t, x; cin >> t >> x;
      S[t].push(x);
    } else if (query == 1) {
      int t; cin >> t;
      if (S[t].size() > 0) cout << S[t].top() << BR;
    } else {
      int t; cin >> t;
      if (S[t].size() > 0) S[t].pop();
    }
  }

  return 0;
}

### Prompt

Develop a solution in Cpp 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 <bits/stdc++.h>
using namespace std;
using ll = long long;

#define rep(i,s,n) for(ll i = (s); i < (n); i++)
#define rep0(i,n) rep(i,0,n)
#define rep1(i,n) rep(i,1,n+1)

#define BR "\n"

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

  stack<int> S[n];

  rep0(i, q) {
    int query; cin >> query;
    if (query == 0) {
      int t, x; cin >> t >> x;
      S[t].push(x);
    } else if (query == 1) {
      int t; cin >> t;
      if (S[t].size() > 0) cout << S[t].top() << BR;
    } else {
      int t; cin >> t;
      if (S[t].size() > 0) S[t].pop();
    }
  }

  return 0;
}

```