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

Solution (in CPP):

Solution:
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <utility>
#include <functional>
#include <string>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
#include <list>
#include <set>
#include <map>
#include <unordered_map>

#define rep(i, n) for(int i = 0; i < (int)(n); i++)
using namespace std;
typedef long long ll;
const int INF = (1<<30);
const int MOD = 1000000007;
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }

int main() {
    int n,Q;
    cin >> n >> Q;
    stack<int> S[n];
    vector<int> ans;
    rep(i, Q){
        int q, t, x;
        cin >> q;
        switch (q)
        {
        case 0:
            cin >> t >> x;
            S[t].push(x);
            break;
        case 1:
            cin >> t;
            if(!S[t].empty()) ans.push_back(S[t].top());
            break;        
        case 2:
            cin >> t;
            if(!S[t].empty()) S[t].pop();
            break;
        }
    }
    int N = ans.size();
    rep(i, N) cout << ans[i] << endl;
    return 0;
}