#include <bits/stdc++.h>
using namespace std;
const int MAXN = 5000000;
int countPrimes(int N) {
    if (N <= 2) {
        return 0; 
    }

    bitset<MAXN> isPrime;
    isPrime.set(); 

    isPrime[0] = isPrime[1] = 0; 
    for (int i = 2; i * i < N; ++i) {
        if (isPrime[i]) {
            for (int j = i * i; j < N; j += i) {
                isPrime[j] = 0;
            }
        }
    }

    int count = 0;
    for (int i = 2; i < N; ++i) {
        if (isPrime[i]) {
            ++count;
        }
    }

    return count;
}

SELECT MAX(months * hackos) AS maximum_hackos, 
       COUNT(*) AS number_of_hackers
FROM HACKER
WHERE months * hackos = (
    SELECT MAX(months * hackos) 
    FROM HACKER
);

WITH combined_views AS (
    SELECT 'Anonymous' AS user_type, id, start_dt, end_dt
    FROM anonymous_viewers
    UNION ALL
    SELECT 'Subscribed' AS user_type, id, start_dt, end_dt
    FROM subscribed_viewers
)
SELECT 
    user_type,
    COUNT(DISTINCT id) AS unique_views,
    COUNT(*) AS total_views
FROM combined_views
GROUP BY user_type
UNION ALL
SELECT 
    'Total' AS user_type,
    COUNT(DISTINCT id) AS unique_views,
    COUNT(*) AS total_views
FROM combined_views
ORDER BY 
    CASE 
        WHEN user_type = 'Anonymous' THEN 1
        WHEN user_type = 'Subscribed' THEN 2
        ELSE 3
    END;

#include <bits/stdc++.h>
using namespace std;

unordered_map<int, string> stringMap;

void addString(int id, char c) {
    stringMap[id] = string(1, c);
}

void concatStrings(int id1, int id2, int id3) {
    stringMap[id3] = stringMap[id1] + stringMap[id2];
    stringMap.erase(id1);
    stringMap.erase(id2);
}

void reverseString(int id) {
    reverse(stringMap[id].begin(), stringMap[id].end());
}

void getCharacter(int id, int k) {
    cout << stringMap[id][k - 1] << endl;
}

void solve() {
    int q;
    cin >> q;

    while (q--) {
        int type;
        cin >> type;

        if (type == 1) {
            int id;
            char c;
            cin >> id >> c;
            addString(id, c);
        } else if (type == 2) {
            int id1, id2, id3;
            cin >> id1 >> id2 >> id3;
            concatStrings(id1, id2, id3);
        } else if (type == 3) {
            int id;
            cin >> id;
            reverseString(id);
        } else if (type == 4) {
            int id, k;
            cin >> id >> k;
            getCharacter(id, k);
        }
    }
}

int main() {
    solve();
    return 0;
}

class KeyHierarchy:
    def __init__(self):
        self.gen = [1] 

    def rotate(self):
        self.gen.append(1)  

    def rekey(self):
        for i in range(len(self.gen)):
            self.gen[i] += 1

    def query(self, fam):
        if 1 <= fam <= len(self.gen):
            return self.gen[fam - 1]
        return 0  

def processOp(operations):
    kh = KeyHierarchy()
    results = []

    for op in operations:
        if op == 'Ro':
            kh.rotate()
        elif op == 'Re':
            kh.rekey()
        elif op.startswith('Q'):
            family = int(op.split()[1])
            results.append(kh.query(family))

    return results

from collections import defaultdict

class BinaryTree:
    def __init__(self, n):
        self.n = n
        self.tree = defaultdict(list)
        self.values = [0] * (n + 1)
    
    def add_edge(self, u, v):
        self.tree[u].append(v)
        self.tree[v].append(u)
    
    def dfs(self, node, parent, depth, max_depth, increment):
        if depth > max_depth:
            return
        
        self.values[node] += increment
        
        for child in self.tree[node]:
            if child != parent:
                self.dfs(child, node, depth + 1, max_depth, increment)
    
    def update(self, ui, di, xi):
        self.dfs(ui, -1, 0, di, xi)

def solve(n, queries):
    tree = BinaryTree(n)
    
    for i in range(2, n + 1):
        parent = i // 2
        tree.add_edge(parent, i)
    
    for ui, di, xi in queries:
        tree.update(ui, di, xi)
    
    return tree.values[1:]

def proBinary(s):
    n = len(s)
    x = 0
    y = 0
    flip = 0
    
    for i in range(n):
        bit = int(s[i])
        x = (x << 1) | bit
        flip = (flip << 1) | (1 - bit)
    
    y = flip
    return f"{x} {y}"