#include <algorithm>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <list>
#include <ostream>
#include <queue>
#include <set>
#include <stack>
#include <vector>
using namespace std;
typedef long long int ll;
#define rep(i, a, n) for (int i = (a); i < (n); i++)
#define repr(i, n, a) for (int i = (n)-1; i >= (a); i--)
#define each(it, v) for (auto(it) = (v).begin(); (it) != (v).end(); (it)++)
#define all(v) (v).begin(), (v).end()
#define vi vector<int>
#define vvi vector<vi>
#define vd vector<double>
#define vvd vector<vd>
#define co(x) cout << (x)
#define con(x) cout << (x) << "\n"
#define cosp(x) cout << (x) << " "
#define out(x) rep(i, 0, x.size()) cout << x[i] << ((i != x.size() - 1) ? " " : "\n")
const double PI = 3.14159265;
const int MAXN = 1 << 15;
const ll LINF = 1LL << 60;
const int INF = 1 << 30;
#define Pr(a) (#a) << "=" << (a) << " "
#define pr(a) cout << Pr(a) << '\n'
///////////////////////////////////////////////////////////

int N;
vector<int> V;
deque<int> D;
list<int> L;

int main() {
  int q;
  cin >> N >> q;
  vector<stack<int>> S(N);
  while (q--) {
    int cmd, t;
    cin >> cmd >> t;
    if (cmd == 0) {
      int x;
      cin >> x;
      S[t].push(x);
    }
    if (cmd == 1) {
      if (!S[t].empty()) cout << S[t].top() << '\n';
    }
    if (cmd == 2) {
      if (!S[t].empty()) S[t].pop();
    }
  }
  return 0;
}

### Prompt

Please provide a cpp coded solution 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 <algorithm>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <list>
#include <ostream>
#include <queue>
#include <set>
#include <stack>
#include <vector>
using namespace std;
typedef long long int ll;
#define rep(i, a, n) for (int i = (a); i < (n); i++)
#define repr(i, n, a) for (int i = (n)-1; i >= (a); i--)
#define each(it, v) for (auto(it) = (v).begin(); (it) != (v).end(); (it)++)
#define all(v) (v).begin(), (v).end()
#define vi vector<int>
#define vvi vector<vi>
#define vd vector<double>
#define vvd vector<vd>
#define co(x) cout << (x)
#define con(x) cout << (x) << "\n"
#define cosp(x) cout << (x) << " "
#define out(x) rep(i, 0, x.size()) cout << x[i] << ((i != x.size() - 1) ? " " : "\n")
const double PI = 3.14159265;
const int MAXN = 1 << 15;
const ll LINF = 1LL << 60;
const int INF = 1 << 30;
#define Pr(a) (#a) << "=" << (a) << " "
#define pr(a) cout << Pr(a) << '\n'
///////////////////////////////////////////////////////////

int N;
vector<int> V;
deque<int> D;
list<int> L;

int main() {
  int q;
  cin >> N >> q;
  vector<stack<int>> S(N);
  while (q--) {
    int cmd, t;
    cin >> cmd >> t;
    if (cmd == 0) {
      int x;
      cin >> x;
      S[t].push(x);
    }
    if (cmd == 1) {
      if (!S[t].empty()) cout << S[t].top() << '\n';
    }
    if (cmd == 2) {
      if (!S[t].empty()) S[t].pop();
    }
  }
  return 0;
}

```