/*
* Flatten the tree using Euler tour, note the end points of subtree of each node in tin[] and tout[].
* Maintain the nodes in each region. Also keep a sorted list of tin[] of the nodes in each region.
* For query (r1, r2), for each node 'i' in r1, binary search for the number of nodes in region r2 with tin[] between [tin[i]+1, tout[i]]. Add them all up to get the answer.
*
* This works within the time limit because of the fact that there can only be a small amount of distinct regions queries (R * (R+1) / 2).
* This means that there are a lot of repeated queries. Sadly, we cannot use a R*R array for keeping track of answers (MLE).
* We use a gp_hash_table to keep track of the queries and their answers. If repeated, we just answer in constant time.
* unordered_map usually missed one test case when I tried, so gets 95.
* gp_hash_table gets 100.
*/
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
#include <bits/stdc++.h>
using namespace std;
#define for_(i, s, e) for (int i = s; i < (int) e; i++)
#define for__(i, s, e) for (ll i = s; i < e; i++)
typedef long long ll;
typedef vector<int> vi;
typedef pair<int, int> ii;
//#define endl '\n'
const int MXN = 200000, MXR = 25000, INF = 1e9+1, k = 100000;
int n, r, q;
int reg[MXN+1], tin[MXN+1], tout[MXN+1], pt = 0;
vi adj[MXN+1];
vi regList[MXR+1], regListId[MXR+1];
gp_hash_table<int, int> seen[MXR+1];
void dfs(int p, int parent) {
regList[reg[p]].push_back(pt);
regListId[reg[p]].push_back(p);
tin[p] = pt++;
for (int i: adj[p]) if (i != parent) dfs(i, p);
tout[p] = pt-1;
}
int main() {
#ifdef shiven
freopen("test.in", "r", stdin);
#endif
ios_base::sync_with_stdio(false);
cin.tie(0);
cin >> n >> r >> q;
cin >> reg[0];
reg[0] -= 1;
for_(i, 1, n) {
int p; cin >> p >> reg[i];
reg[i] -= 1;
adj[p-1].push_back(i);
}
dfs(0, 0);
while (q--) {
int r1, r2; cin >> r1 >> r2;
r1 -= 1; r2 -= 1;
if (seen[r1].find(r2) != seen[r1].end()) {
cout << seen[r1][r2] << endl;
continue;
}
int ans = 0;
for (auto i: regListId[r1])
ans += upper_bound(regList[r2].begin(), regList[r2].end(), tout[i]) - lower_bound(regList[r2].begin(), regList[r2].end(), tin[i]+1);
cout << ans << endl;
seen[r1][r2] = ans;
if (seen[r1].size() >= k) seen[r1].clear();
}
return 0;
}
