#pragma GCC optimize("O3","inline")
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
static const int MAXL = 33; // we’ll store bits 32..0
struct Node {
Node* a[2] = {nullptr,nullptr};
set<int> times;
};
struct Trie {
Node* head;
Trie() { head = new Node(); }
};
struct Query {
bool isAdd;
int x; // for Add: parent‑index, for Query: node a
int y; // for Add: weight , for Query: node b
};
int main(){
ios::sync_with_stdio(false);
cin.tie(nullptr);
int Q;
cin >> Q;
vector<Query> qs(Q);
int totalNodes = 1;
// 1) read queries and count how many nodes we'll have
for(int i = 0; i < Q; i++){
string ty;
int x; ll w;
cin >> ty >> x >> w;
if (ty == "Add") {
qs[i] = { true, x-1, (int)w };
totalNodes++;
} else {
// Query a b
qs[i] = { false, x-1, (int)(w-1) };
}
}
// 2) build the static tree, do one DFS to get xra[], tin[], tout[]
vector<vector<pair<int,ll>>> adj(totalNodes);
int nextId = 1;
for(auto &q : qs){
if (q.isAdd) {
adj[q.x].emplace_back(nextId++, q.y);
}
}
vector<ll> xra(totalNodes, 0);
vector<int> tin(totalNodes), tout(totalNodes);
int timer = 0;
function<void(int)> dfs = [&](int u){
tin[u] = timer++;
for(auto &e : adj[u]){
int v = e.first;
ll w = e.second;
xra[v] = xra[u] ^ w;
dfs(v);
}
tout[u] = timer++;
};
dfs(0);
// 3) second pass: interleave trie‑inserts on Add and greedy XOR‑queries
Trie trie;
vector<ll> answers;
nextId = 1; // for Add ordering
for(auto &q : qs){
if (q.isAdd) {
int u = nextId++;
ll val = xra[u];
Node* cur = trie.head;
for(int b = MAXL-1; b >= 0; --b){
int bit = (val >> b) & 1;
if (!cur->a[bit])
cur->a[bit] = new Node();
cur = cur->a[bit];
cur->times.insert(tin[u]);
}
} else {
int a = q.x, b = q.y;
ll X = xra[a];
int L = tin[b], R = tout[b];
ll ans = 0;
Node* cur = trie.head;
for(int bbit = MAXL-1; bbit >= 0; --bbit){
int want = 1 - ((X >> bbit) & 1);
Node* cand = cur->a[want];
bool ok = false;
if (cand) {
auto it = cand->times.lower_bound(L);
if (it != cand->times.end() && *it <= R)
ok = true;
}
if (ok) {
ans |= (1LL << bbit);
cur = cand;
} else {
cur = cur->a[1-want];
}
}
answers.push_back(ans);
}
}
// 4) output answers
for (ll v : answers)
cout << v << " ";
cout << "\n";
return 0;
}
