#include <bits/stdc++.h>
#define REP(i, n) for (int i = 0; i < (int)(n); i++)
using namespace std;
template<class T>inline bool chmax(T &a, const T &b){if(a < b){a = b; return 1;}return 0;}
template<class T>inline bool chmin(T &a, const T &b){if(a > b){a = b; return 1;}return 0;}
typedef long long ll;
int main(){
  ios::sync_with_stdio(false);
  cin.tie(0);
  int N, Q; cin >> N >> Q;
  vector<stack<int> > S(N, stack<int>());
  vector<int> ans;
  REP(i, Q){
    int q, t, x; cin >> q >> t;
    if(q == 0){
      cin >> x;
      S[t].push(x);
    }else if(q == 1){
      if(!S[t].empty()) ans.push_back(S[t].top());
    }else{
      if(!S[t].empty()) S[t].pop();
    }
  }
  for(auto &i : ans) cout << i << endl;
}


### Prompt

Create a solution in Cpp for 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 <bits/stdc++.h>
#define REP(i, n) for (int i = 0; i < (int)(n); i++)
using namespace std;
template<class T>inline bool chmax(T &a, const T &b){if(a < b){a = b; return 1;}return 0;}
template<class T>inline bool chmin(T &a, const T &b){if(a > b){a = b; return 1;}return 0;}
typedef long long ll;
int main(){
  ios::sync_with_stdio(false);
  cin.tie(0);
  int N, Q; cin >> N >> Q;
  vector<stack<int> > S(N, stack<int>());
  vector<int> ans;
  REP(i, Q){
    int q, t, x; cin >> q >> t;
    if(q == 0){
      cin >> x;
      S[t].push(x);
    }else if(q == 1){
      if(!S[t].empty()) ans.push_back(S[t].top());
    }else{
      if(!S[t].empty()) S[t].pop();
    }
  }
  for(auto &i : ans) cout << i << endl;
}


```