#include <bits/stdc++.h>
#define all(x) (x).begin(), (x).end()
#ifndef ONLINE_JUDGE
template<typename T>
void pr(T a){std::cerr<<a<<std::endl;}
template<typename T,typename... Args>
void pr(T a, Args... args) {std::cerr<<a<<' ',pr(args...);}
#else
template<typename... Args>
void pr(Args... args){}
#endif
using namespace std;
const int MM = 2e5+5, SQ = 500;
int n, m, q;
vector<int> adj[MM];
int in[MM], out[MM], t;
map<int, int> cnt[MM], parcnt;
int val[MM];
vector<pair<int, int>> down[MM], up[MM];
vector<int> nodes[MM];
int ans[MM], bit[MM];
void update(int i, int v){
for(; i < MM; i += i&-i)
bit[i] += v;
}
int query(int i){
int v = 0;
for(; i; i -= i&-i)
v += bit[i];
return v;
}
void dfs(int cur, int pre){
in[cur] = ++t;
for(auto u: adj[cur]){
if(u == pre) continue;
dfs(u, cur);
}
out[cur] = t;
}
void go(int cur, int pre){
parcnt[val[cur]]++;
for(auto [a, i]: up[cur]){
ans[i] += parcnt[a];
}
int id = 0;
for(auto u: adj[cur]){
if(u == pre) continue;
go(u, cur);
if(size(cnt[u]) > size(cnt[id]))
id = u;
}
cnt[cur] = move(cnt[id]);
for(auto u: adj[cur]){
if(u == pre) continue;
if(u != id){
for(auto [a, c]: cnt[u]){
cnt[cur][a] += c;
}
}
}
cnt[cur][val[cur]]++;
for(auto [a, i]: down[cur]){
ans[i] += cnt[cur][a];
}
parcnt[val[cur]]--;
}
int main(){
ios_base::sync_with_stdio(0);
cin.tie(0);
cin>>n>>m>>q;
cin>>val[1];
nodes[val[1]].emplace_back(1);
for(int i = 2,p; i <= n; i++){
cin>>p>>val[i];
// pr(p, i, val[i]);
adj[p].emplace_back(i);
nodes[val[i]].emplace_back(i);
}
dfs(1, 0);
for(int i = 0,a,b; i < q; i++){
cin>>a>>b;
// if(size(nodes[a]) > SQ){
// for(int j: nodes[b])
// up[j].emplace_back(a, i);
// }
// else if(size(nodes[b]) > SQ){
// for(int j: nodes[a])
// down[j].emplace_back(b, i);
// }
// else
{
for(int j: nodes[b]){
update(in[j], 1);
}
for(int j: nodes[a]){
ans[i] += query(out[j]) - query(in[j]-1);
}
for(int j: nodes[b]){
update(in[j], -1);
}
}
}
// go(1, 0);
for(int i = 0; i < q; i++)
cout<<ans[i]<<'\n';
}
/*
for pairs < sqrt, update and query on ett O(n log n)
dfs small to large
O(n log n) for inserting
to query, use small to query for large
O(small*large)
O(n sqrt n)
large to large
O(sqrtn sqrtn) = O(n)
so during dfs, I need one map to store the set of parents and another for subtree (this needs small to large)
*/
