#include <bits/stdc++.h>
using namespace std;
using ll = long long;
template<class T> bool cmin(T &i, T j) { return i > j ? i=j,true:false; }
template<class T> bool cmax(T &i, T j) { return i < j ? i=j,true:false; }
const int nax = 2e5, rax = 25000;
int st[4*nax];
int n;
vector<int> adj[nax];
vector<int> nodes[rax];
int tin[nax],tout[nax],color[nax],timer=0;
map<pair<int,int>,ll> computed;
void upd(int p,int nw,int u=1,int tl=0,int tr=n-1) {
if (tl == tr) {
st[u] = nw;
return;
}
int mid = (tl+tr)/2;
if (p <= mid) {
upd(p,nw,2*u,tl,mid);
} else {
upd(p,nw,2*u+1,mid+1,tr);
}
st[u] = st[2*u] + st[2*u+1];
}
int qry(int l,int r,int u=1,int tl=0,int tr=n-1) {
if (l > r) {
return 0;
}
if (l == tl && r == tr) {
return st[u];
}
int mid = (tl+tr)/2;
return qry(l,min(r,mid),2*u,tl,mid)+qry(max(mid+1,l),r,2*u+1,mid+1,tr);
}
void dfs(int u=0,int e=-1) {
tin[u] = timer++;
for (int &v: adj[u]) if (v != e) {
dfs(v,u);
}
tout[u] = timer-1;
}
void solve() {
int r,q;
cin >> n >> r >> q;
cin >> color[0];
for (int i=1,p;i<n;i++) {
cin >> p >> color[i];
--p;
adj[p].push_back(i);
adj[i].push_back(p);
}
for (int i=0;i<n;i++) {
color[i]--;
}
dfs();
for (int i=0;i<n;i++) {
nodes[color[i]].push_back(i);
}
for (int a,b;q--;) {
cin >> a >> b;
--a,--b;
auto c = make_pair(a,b);
if (computed.count(c)) { cout << computed[c] << endl; continue; }
ll ans = 0;
// cout << "\nSon-nodes:\n";
for (auto &x: nodes[b]) {
// cout << x << " ";
upd(tin[x],1);
}
// cout << "\nFather nodes:\n";
for (auto &x: nodes[a]) {
// cout << x << " ";
ans += qry(tin[x],tout[x]);
}
for (auto &x: nodes[b]) {
upd(tin[x],0);
}
computed[c] = ans;
cout << ans << endl;
}
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
int _t = 1;
for (int i=1;i<=_t;i++) {
solve();
}
}
// Dato un albero in cui ogni nodo ha un tag, query online che chiedono il numero di coppie di nodi con tag E1 ed E2, tali che
// E1 e' un antenato di E2.
