// IOI2009 – Regions
#include <bits/stdc++.h>
using namespace std;
using VI = vector<int>;
using IP = pair<int, int>;
using LL = long long;
struct Emp { // Employee
int tin, reg; // tin(dfs时间戳), reg(区域)
VI ch; // 被指导的
};
struct Reg {
VI ids; // 其中的Employee
/* Sorted of intervals with the same nesting level. Each pair is
* (ID, depth) where ID is the left end-point of the interval (inclusive).
* The right end-point is implicit from the following interval.
*/
vector<IP> ranges;
int depth; /* Working depth during the DFS. */
};
vector<Emp> ES;
vector<Reg> RS;
/* Does a pre-order walk over the subtree rooted at root. id_pool contains
* the next unused employee ID, and on return it will be updated to again
* be the next available ID.
* This procedure builds the RS arrays, after which the tree is no longer needed.
*/
void dfs(int u, int &timer) {
auto& r = RS[ES[u].reg];
r.ids.push_back(timer++);
/* Depth changed, so after this point we need a new range */
r.ranges.push_back({timer, ++r.depth});
for (int v : ES[u].ch)
dfs(v, timer);
/* Undo the depth change, and start another interval after the last managee. */
r.ranges.push_back({timer, --r.depth});
}
/* Query in O(R2 log R1) time, by counting for each employee in r2. */
LL query_by_id(const Reg &r1, const Reg &r2) {
LL ans = 0;
for (int pos : r2.ids) {
/* Find the first range that starts at pos or later. This will
* actually be the range after the one we want.
*/
auto it = lower_bound(r1.ranges.begin(), r1.ranges.end(), make_pair(pos, INT_MAX));
if (it != r1.ranges.begin())
ans += prev(it)->second;
}
return ans;
}
/* Query in O(R1 log R2) time, by counting for each employee in r1 */
LL query_by_range(const Reg &r1, const Reg &r2) {
LL ans = 0;
for (size_t i = 0; i + 1 < r1.ranges.size(); i++) {
int p1 = r1.ranges[i].first, p2 = r1.ranges[i + 1].first;
/* Each employee from r2 in [p1, p2) has depth managers
* from r1. Find the intersections of [p1, p2) with the
* employee list for r2.
*/
auto it1 = lower_bound(r2.ids.begin(), r2.ids.end(), p1),
it2 = lower_bound(r2.ids.begin(), r2.ids.end(), p2);
ans += r1.ranges[i].second * (it2 - it1);
}
return ans;
}
/* Query in O(R1 + R2) time, by counting for each employee in r1
* but with a linear sweep instead of a binary search.
*/
LL query_stitch(const Reg &r1, const Reg &r2) {
LL ans = 0;
/* Find the first employee id that is in the first range */
auto id = r2.ids.begin();
if (r1.ranges.empty()) return ans;
while (id != r2.ids.end() && *id < r1.ranges[0].first) id++;
/* Iterate over the ranges as above */
for (size_t i = 0; i + 1 < r1.ranges.size() && id != r2.ids.end(); i++) {
/* Find the end of the section of employees from this range */
auto id_bak = id;
while (id != r2.ids.end() && *id < r1.ranges[i + 1].first)
id++;
ans += r1.ranges[i].second * (id - id_bak);
}
return ans;
}
LL solve(int r1, int r2) {
static map<IP, LL> cache;
IP key(r1, r2);
if (cache.count(key)) return cache[key];
const Reg ®1 = RS[r1], ®2 = RS[r2];
int sz1 = reg1.ids.size(), sz2 = reg2.ids.size();
int costs[3] = {
sz1 * ((int)log2(sz2) + 2) * 5, sz2 * ((int)log2(sz1) + 2) * 5, sz1 + sz2
};
int k = min_element(costs, costs + 3) - costs;
if (k == 0) return cache[key] = query_by_range(reg1, reg2);
if (k == 1) return cache[key] = query_by_id(reg1, reg2);
return cache[key] = query_stitch(reg1, reg2);
}
int main() {
ios::sync_with_stdio(false), cin.tie(0);
int N, R, Q, timer = 0;
cin >> N >> R >> Q, ES.resize(N), RS.resize(R);
cin >> ES[0].reg, --ES[0].reg;
for (int i = 1, fa; i < N; i++)
cin >> fa >> ES[i].reg, --ES[i].reg, ES[fa - 1].ch.push_back(i);
dfs(0, timer);
for (int q = 0, r1, r2; q < Q; q++)
cin >> r1 >> r2, cout << solve(r1 - 1, r2 - 1) << endl;
return 0;
}
