#include <iostream>
#include <vector>

using namespace std;

struct Stack {
    int pos = -1;
    vector<int> data;
};

void push(Stack &t, int x) {
    if (t.pos == t.data.size() - 1) {
        t.data.push_back(x);
        t.pos++;
    } else
        t.data[++t.pos] = x;
}

void top(Stack &t) {
    if (t.pos > -1)
        cout << t.data[t.pos] << endl;
}

void pop(Stack &t) {
    if (t.pos > -1)
        t.pos--;
}

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

    Stack S[n];

    while (Q--) {
        int q, t;
        cin >> q >> t;
        switch (q) {
            case 0:
                int x;
                cin >> x;
                push(S[t], x);
                break;
            case 1:
                top(S[t]);
                break;
            case 2:
                pop(S[t]);
                break;
        }
    }
}

### 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 <iostream>
#include <vector>

using namespace std;

struct Stack {
    int pos = -1;
    vector<int> data;
};

void push(Stack &t, int x) {
    if (t.pos == t.data.size() - 1) {
        t.data.push_back(x);
        t.pos++;
    } else
        t.data[++t.pos] = x;
}

void top(Stack &t) {
    if (t.pos > -1)
        cout << t.data[t.pos] << endl;
}

void pop(Stack &t) {
    if (t.pos > -1)
        t.pos--;
}

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

    Stack S[n];

    while (Q--) {
        int q, t;
        cin >> q >> t;
        switch (q) {
            case 0:
                int x;
                cin >> x;
                push(S[t], x);
                break;
            case 1:
                top(S[t]);
                break;
            case 2:
                pop(S[t]);
                break;
        }
    }
}

```