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 <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll INF = 1e18;
const ll mod = 1000000007;

#define rep(i, n) for (int i = 0; i < (int)(n); i++)
template<class T> inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template<class T> inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
int ctoi(char c) {
  if (c >= '0' && c <= '9') {
    return c - '0';
  }
  return 0;
}
ll factorial(ll n) {
  if (n == 1) {
    return 1;
  }
  ll ans = factorial(n-1);
  return ans*n;
}
ll gcd(ll a, ll b) {
  if (a < b) {
    ll tmp = a;
    a = b;
    b = tmp;
  }
  ll r = a%b;
  while(r != 0) {
    a = b;
    b = r;
    r = a%b;
  }
  return b;
}
ll lcm(ll a, ll b) {
  return (a/gcd(a, b))*b;
}

int main() {
  ll n, m;
  cin >> n >> m;
  vector <stack <ll>> v(n, stack <ll> ());
  rep(i, m) {
    ll a, b, c;
    cin  >> a >> b;
    if (a == 0) {
      cin >> c;
      v.at(b).push(c);
    } else if (a == 1) {
      if (!v.at(b).empty()) {
        cout << v.at(b).top() << endl;
      }
    } else {
      if (!v.at(b).empty()) {
        v.at(b).pop();
      }
    }
  }
}