/**
* Solution by 1egend (polarity.sh)
* Date: 2025-02-15
* Contest: IOI 2009
* Problem: regions
**/
#include <bits/stdc++.h>
using namespace std;
using ull = unsigned long long;
using ll = long long;
using vi = vector<int>;
using vl = vector<ll>;
using pii = pair<int, int>;
#define pb push_back
#define rep(i, a, b) for(int i = (a); i < (b); ++i)
#define all(x) begin(x), end(x)
#define sz(x) (int)(x).size()
const int MAX_N = 1e5 + 1;
const ll MOD = 1e9 + 7;
/**
* ALGORITHM: Euler Tour
* PURPOSE: Flattens a tree so that each range contains any subtree range
* CONSTRAINT: Graph must be a tree
* TIME: O(V)
*/
vector<pii> eulerTour(int n, vector<vector<int>> &adj){
vector<pii> ans(n);
int i = -1;
function<void(int, int)> tour;
tour = [&](int node, int parent){
ans[node].first = ++i;
for (int x : adj[node]){
if (x != parent){
tour(x, node);
}
}
ans[node].second = i;
};
// root at 0
tour(0, 0);
return ans;
}
void solve(){
int n, r, q; cin >> n >> r >> q;
vector<vi> adj(n);
vi region(n);
rep(i, 0, n){
if (i == 0){
cin >> region[0];
region[0]--;
continue;
}
int s; cin >> s;
--s;
adj[i].pb(s);
adj[s].pb(i);
cin >> region[i];
region[i]--;
}
vector<pii> tree = eulerTour(n, adj);
vi rend(n);
vi arr(n);
rep(i, 0, n){
arr[tree[i].first] = region[i];
rend[tree[i].first] = tree[i].second;
}
vi curr(r, 0);
vector<vi> ans(r, vi(r, 0));
using T = pii;
priority_queue<T, vector<T>, greater<T>> pq;
rep(i, 0, n){
int a = arr[i];
pq.push(pii{rend[i], a});
curr[a]++;
while(pq.top().first < i){
curr[pq.top().second]--;
pq.pop();
}
rep(j, 0, r){
ans[j][a] += curr[j];
}
}
rep(i, 0, q){
int r1, r2; cin >> r1 >> r2;
--r1; --r2;
cout << ans[r1][r2] << endl;
}
}
int main(){
ios_base::sync_with_stdio(0);
cin.tie(0); cout.tie(0);
solve();
return 0;
}
