#include <bits/stdc++.h>
using namespace std;
struct Fast {
    Fast() {
        cin.tie(0);
        ios::sync_with_stdio(false);
        cout << fixed << setprecision(20);
    }
} fast;

long long mod = 998244353;
#define REP(i, n) for (long long i = 0; i < (n); i++)

long long modpow(long long m, long long n) {
    if (n == 0) return 1;
    if (n % 2 == 0) {
        long long t = modpow(m, n / 2);
        return t * t;
    } else {
        return modpow(m, n - 1) * m;
    }
}

long long moddiv(long long a, long long b) { return (a * modpow(b, mod - 2)) % mod; }

long long nCkmod(long long n, long long k) {
    long long denomi = 1;
    REP(i, n) {
        denomi *= (i + 1);
        denomi %= mod;
    }
    long long numera = 1;
    for (int i = n - k; i > 0; i--) {
        numera *= i;
        numera %= mod;
    }
    REP(i, k) {
        numera *= (i + 1);
        numera %= mod;
    }
    return moddiv(denomi, numera);
}

void print_binary(long long a) {
    for (int i = 31; i >= 0; i--) {
        cout << (a >> i & 1);
    }
    cout << endl;
}

void yes() {
    cout << "Yes" << endl;
    exit(0);
}
void no() {
    cout << "No" << endl;
    exit(0);
}
array<int, 101> factoring(int x) {
    array<int, 101> ret{};
    for (int i = 2; x >= i * i; i++) {
        while (x % i == 0) {
            ret[i]++;
            x /= i;
        }
    }
    if (x > 1) ret[x]++;
    return ret;
}

signed main() {
    int n, q;
    cin >> n >> q;
    vector<stack<int>> st(n);
    REP(i, q) {
        int a, t, x;
        cin >> a >> t;
        if (a == 0) {
            cin >> x;
            st[t].push(x);
        } else if (a == 1 && st[t].size() != 0) {
            cout << st[t].top() << endl;
        } else if (st[t].size() != 0) {
            st[t].pop();
        }
    }
}

### Prompt

Generate 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;
struct Fast {
    Fast() {
        cin.tie(0);
        ios::sync_with_stdio(false);
        cout << fixed << setprecision(20);
    }
} fast;

long long mod = 998244353;
#define REP(i, n) for (long long i = 0; i < (n); i++)

long long modpow(long long m, long long n) {
    if (n == 0) return 1;
    if (n % 2 == 0) {
        long long t = modpow(m, n / 2);
        return t * t;
    } else {
        return modpow(m, n - 1) * m;
    }
}

long long moddiv(long long a, long long b) { return (a * modpow(b, mod - 2)) % mod; }

long long nCkmod(long long n, long long k) {
    long long denomi = 1;
    REP(i, n) {
        denomi *= (i + 1);
        denomi %= mod;
    }
    long long numera = 1;
    for (int i = n - k; i > 0; i--) {
        numera *= i;
        numera %= mod;
    }
    REP(i, k) {
        numera *= (i + 1);
        numera %= mod;
    }
    return moddiv(denomi, numera);
}

void print_binary(long long a) {
    for (int i = 31; i >= 0; i--) {
        cout << (a >> i & 1);
    }
    cout << endl;
}

void yes() {
    cout << "Yes" << endl;
    exit(0);
}
void no() {
    cout << "No" << endl;
    exit(0);
}
array<int, 101> factoring(int x) {
    array<int, 101> ret{};
    for (int i = 2; x >= i * i; i++) {
        while (x % i == 0) {
            ret[i]++;
            x /= i;
        }
    }
    if (x > 1) ret[x]++;
    return ret;
}

signed main() {
    int n, q;
    cin >> n >> q;
    vector<stack<int>> st(n);
    REP(i, q) {
        int a, t, x;
        cin >> a >> t;
        if (a == 0) {
            cin >> x;
            st[t].push(x);
        } else if (a == 1 && st[t].size() != 0) {
            cout << st[t].top() << endl;
        } else if (st[t].size() != 0) {
            st[t].pop();
        }
    }
}

```