#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
// const ll M = 998244353;
const ll M = 1e9 + 7;
#define all(v) (v).begin(), (v).end()
#define ff first
#define ss second
const ll INFL = 1e18;
const int INF = 1e9;
const int mxN = 2e5;
const int mxR = 25000;
const int Z = 2000;
vector<int> adj[mxN];
int freq[mxR], tin[mxN], tout[mxN], ord[mxR], color[mxN], rnk[mxR];
int pre[Z][Z];
int cur[Z];
int timer = -1;
int n, r, q;
vector<int> with_color[mxR];
vector<int> tins[mxR];
vector<array<int,2 >> edge_pref[mxR];
void dfs(int v) {
tin[v] = ++timer;
edge_pref[color[v]].push_back({tin[v], edge_pref[color[v]].back()[1]+1});
tins[color[v]].push_back(tin[v]);
if (rnk[color[v]]<Z) {
for (int i=0; i<Z; i++) {
pre[i][rnk[color[v]]] += cur[i];
}
cur[rnk[color[v]]]++;
}
for (auto u: adj[v]) dfs(u);
if (rnk[color[v]]<Z) cur[rnk[color[v]]]--;
tout[v] = ++timer;
edge_pref[color[v]].push_back({tout[v], edge_pref[color[v]].back()[1]-1});
}
// for each in a calc count of b in subtree
int calc_1(int a, int b) {
int res = 0;
for (auto i: with_color[a]) {
res += upper_bound(all(tins[b]), tout[i])-lower_bound(all(tins[b]), tin[i]);
}
return res;
}
// for each in b calc parents with a
int calc_2(int a, int b) {
int res = 0;
for (auto i: with_color[b]) {
auto it = upper_bound(all(edge_pref[a]), array<int,2>{tin[i], 0});
it--;
res += (*it)[1];
}
return res;
}
void solve() {
cin >> n >> r >> q;
cin >> color[0];
color[0]--;
for (int i=1; i<n; i++) {
int s;
cin >> s >> color[i];
color[i]--; s--;
adj[s].push_back(i);
}
for (int i=0; i<n; i++) freq[color[i]]++;
iota(ord, ord+r, 0);
sort(ord, ord+r, [&](int l, int r) {
return freq[l] > freq[r];
});
for (int i=0; i<r; i++) rnk[ord[i]] = i;
for (int i=0; i<r; i++) edge_pref[i].push_back({-1, 0});
for (int i=0; i<n; i++) with_color[color[i]].push_back(i);
dfs(0);
while (q--) {
int r1, r2;
cin >> r1 >> r2;
r1--; r2--;
cout << calc_2(r1, r2) << '\n';
// if (rnk[r1]>=Z) cout << calc_1(r1, r2) << '\n';
// else if (rnk[r2]>=Z) cout << calc_2(r1, r2) << '\n';
// else cout << pre[r1][r2] << '\n';
}
}
int main() {
cin.tie(0)->sync_with_stdio(0);
int tc=1;
// cin >> tc;
while (tc--) solve();
}
