#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <utility>
#include <stack>
#include <numeric>
#include <algorithm>
#include <stdio.h>
#include <bitset>
#include <limits.h>
#include <complex>
#include <iomanip>
using namespace std;
#define MM 1000000000
#define MOD MM+7
#define pi pair<int32,int32>
#define pl pair<int64,int64>
#define chmax(a,b) (a<b ? a=b:0)
#define chmin(a,b) (a>b ? a=b:0)
const long double PI = acos(-1);
const long long INF = 1e15;
int dx[8] = {-1,0,1,0,-1,-1,1,1};
int dy[8] = {0,-1,0,1,-1,1,1,-1};
template<typename T>
T GCD(T u,T v){
    return v ? GCD(v,u%v) : u;
}
int main(){
    int n,q; cin >> n >> q;
    stack<int> st[n];
    for(int i = 0; i < q; i++){
        int num; cin >> num;
        if (num == 0){
            int t,x; cin >> t >> x;
            st[t].push(x);
        } else if (num == 1){
            int t; cin >> t;
            if (st[t].size()){
                cout << st[t].top() << endl;
            }
        } else if (num == 2){
            int t; cin >> t;
            if (st[t].size()){
                st[t].pop();
            }
        }
    }
}
  



### Prompt

In CPP, your task is to solve 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 <iostream>
#include <string>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <utility>
#include <stack>
#include <numeric>
#include <algorithm>
#include <stdio.h>
#include <bitset>
#include <limits.h>
#include <complex>
#include <iomanip>
using namespace std;
#define MM 1000000000
#define MOD MM+7
#define pi pair<int32,int32>
#define pl pair<int64,int64>
#define chmax(a,b) (a<b ? a=b:0)
#define chmin(a,b) (a>b ? a=b:0)
const long double PI = acos(-1);
const long long INF = 1e15;
int dx[8] = {-1,0,1,0,-1,-1,1,1};
int dy[8] = {0,-1,0,1,-1,1,1,-1};
template<typename T>
T GCD(T u,T v){
    return v ? GCD(v,u%v) : u;
}
int main(){
    int n,q; cin >> n >> q;
    stack<int> st[n];
    for(int i = 0; i < q; i++){
        int num; cin >> num;
        if (num == 0){
            int t,x; cin >> t >> x;
            st[t].push(x);
        } else if (num == 1){
            int t; cin >> t;
            if (st[t].size()){
                cout << st[t].top() << endl;
            }
        } else if (num == 2){
            int t; cin >> t;
            if (st[t].size()){
                st[t].pop();
            }
        }
    }
}
  



```