#include <bits/stdc++.h>

using namespace std;

typedef long long int ll;
typedef pair<ll, ll> pll;

#define FOR(i, n, m) for(ll (i)=(m);(i)<(n);++(i))
#define REP(i, n) FOR(i,n,0)
#define OF64 std::setprecision(10)

const ll MOD = 1000000007;
const ll INF = (ll) 1e15;

ll S[1005][200005];
ll I[1005];

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    ll N, Q;
    cin >> N >> Q;
    memset(I, 0, sizeof(I));
    REP(_, Q) {
        ll q;
        cin >> q;
        if (q == 0) {
            ll t, x;
            cin >> t >> x;
            S[t][I[t]++] = x;
        }
        else if (q == 1) {
            ll t;
            cin >> t;
            if (I[t] == 0)
                continue;
            cout << S[t][I[t] - 1] << endl;
        }
        else {
            ll t;
            cin >> t;
            if (I[t] == 0)
                continue;
            I[t]--;
        }
    }
    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;

typedef long long int ll;
typedef pair<ll, ll> pll;

#define FOR(i, n, m) for(ll (i)=(m);(i)<(n);++(i))
#define REP(i, n) FOR(i,n,0)
#define OF64 std::setprecision(10)

const ll MOD = 1000000007;
const ll INF = (ll) 1e15;

ll S[1005][200005];
ll I[1005];

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    ll N, Q;
    cin >> N >> Q;
    memset(I, 0, sizeof(I));
    REP(_, Q) {
        ll q;
        cin >> q;
        if (q == 0) {
            ll t, x;
            cin >> t >> x;
            S[t][I[t]++] = x;
        }
        else if (q == 1) {
            ll t;
            cin >> t;
            if (I[t] == 0)
                continue;
            cout << S[t][I[t] - 1] << endl;
        }
        else {
            ll t;
            cin >> t;
            if (I[t] == 0)
                continue;
            I[t]--;
        }
    }
    return 0;
}

```