#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0;i<n;i++)
#define REP(i,s,n) for(int i=s;i<n;i++)
#define all(a) a.begin(),a.end()
typedef long long ll;
const ll inf = 1e9;
int main(){
    int n, q;
    cin >> n >> q;
    vector<stack<ll> > vs(n);
    rep(i,q){
        ll com, t; 
        cin >> com >> t;
        ll x;
        if(com == 0){
            cin >> x;
            vs[t].push(x);
        }else if(com == 1 && vs[t].size() > 0)cout << vs[t].top() << endl;
        else if(vs[t].size() > 0)vs[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 <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0;i<n;i++)
#define REP(i,s,n) for(int i=s;i<n;i++)
#define all(a) a.begin(),a.end()
typedef long long ll;
const ll inf = 1e9;
int main(){
    int n, q;
    cin >> n >> q;
    vector<stack<ll> > vs(n);
    rep(i,q){
        ll com, t; 
        cin >> com >> t;
        ll x;
        if(com == 0){
            cin >> x;
            vs[t].push(x);
        }else if(com == 1 && vs[t].size() > 0)cout << vs[t].top() << endl;
        else if(vs[t].size() > 0)vs[t].pop();
    }
    return 0;
} 

```