#include <bits/stdc++.h>
#define ull unsigned long long
#define ll long long
#define il inline
#define db double
#define ls rt << 1
#define rs rt << 1 | 1
#define pb push_back
#define mp make_pair
#define pii pair<int, int>
#define pll pair<ll, ll>
#define X first
#define Y second
#define pcc pair<char, char>
#define vi vector<int>
#define vl vector<ll>
#define rep(i, x, y) for(int i = x; i <= y; i ++)
#define rrep(i, x, y) for(int i = x; i >= y; i --)
#define rep0(i, n) for(int i = 0; i < (n); i ++)
#define per0(i, n) for(int i = (n) - 1; i >= 0; i --)
#define ept 1e-9
#define INF 0x3f3f3f3f
#define sz(x) (x).size()
#define All(x) (x).begin(), (x).end()
#define lowbit(x) ((x) & -(x))
using namespace std;
inline int read()
{
	int x = 0, f = 1; char ch = getchar();
	while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar(); }
	while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
	return x * f;
}
inline ll read1()
{
	ll x = 0, f = 1; char ch = getchar();
	while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar(); }
	while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
	return x * f;
}
const ll mod = 1e9 + 7;
const int N = 4e6 + 10;
const db pi = acos(-1);
stack<int> a[1010];
int main()
{
	int n = read();
	int Q = read();
	
	while(Q --)
	{
		int op = read();
		if(op == 0) {int x = read();a[x].push(read());}
		if(op == 1) 
		{
			int x = read();
			if(a[x].size() != 0) printf("%d\n", a[x].top());
		}
		if(op == 2) 
		{
			int x = read();
			if(a[x].size() != 0) a[x].pop();
		}
	}
	return 0;
}


### Prompt

Your task is to create 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>
#define ull unsigned long long
#define ll long long
#define il inline
#define db double
#define ls rt << 1
#define rs rt << 1 | 1
#define pb push_back
#define mp make_pair
#define pii pair<int, int>
#define pll pair<ll, ll>
#define X first
#define Y second
#define pcc pair<char, char>
#define vi vector<int>
#define vl vector<ll>
#define rep(i, x, y) for(int i = x; i <= y; i ++)
#define rrep(i, x, y) for(int i = x; i >= y; i --)
#define rep0(i, n) for(int i = 0; i < (n); i ++)
#define per0(i, n) for(int i = (n) - 1; i >= 0; i --)
#define ept 1e-9
#define INF 0x3f3f3f3f
#define sz(x) (x).size()
#define All(x) (x).begin(), (x).end()
#define lowbit(x) ((x) & -(x))
using namespace std;
inline int read()
{
	int x = 0, f = 1; char ch = getchar();
	while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar(); }
	while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
	return x * f;
}
inline ll read1()
{
	ll x = 0, f = 1; char ch = getchar();
	while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar(); }
	while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
	return x * f;
}
const ll mod = 1e9 + 7;
const int N = 4e6 + 10;
const db pi = acos(-1);
stack<int> a[1010];
int main()
{
	int n = read();
	int Q = read();
	
	while(Q --)
	{
		int op = read();
		if(op == 0) {int x = read();a[x].push(read());}
		if(op == 1) 
		{
			int x = read();
			if(a[x].size() != 0) printf("%d\n", a[x].top());
		}
		if(op == 2) 
		{
			int x = read();
			if(a[x].size() != 0) a[x].pop();
		}
	}
	return 0;
}


```