#include <bits/stdc++.h>
using namespace std;
#define task "REGIONS"
#define FOR(i, a, b) for (int i = (a); i <= (b); i++)
#define FOD(i, a, b) for (int i = (a); i >= (b); i--)
#define F first
#define S second
const int N = (int)2e5 + 5;
const int BLOCK = 450;
int numNode, numQuery, numRegions;
vector<int> adj[N];
vector<pair<int, int>> regions[25005];
int region[N];
int timer = 0;
int in[N], out[N];
void dfs(int u, int p) {
in[u] = ++timer;
for (int v : adj[u]) if (v != p)
dfs(v, u);
out[u] = timer;
}
struct NODE {
int u, region, timer;
bool operator < (const NODE &other) const {
return timer < other.timer;
}
};
vector<NODE> nodes;
int ans[BLOCK][25005];
int id[25005];
void solve(int r1, int r2) {
int res = 0;
for (auto [t, u] : regions[r1]) {
if (u == 0) break;
int l = lower_bound(regions[r2].begin(), regions[r2].end(), make_pair(t, N)) - regions[r2].begin();
int r = upper_bound(regions[r2].begin(), regions[r2].end(), make_pair(out[u], N)) - regions[r2].begin() - 1;
res += max(0, r - l + 1);
}
cout << res << "\n";
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0); cout.tie(0);
cin >> numNode >> numRegions >> numQuery;
FOR(i, 1, numNode) {
int p = -1;
if (i == 1) cin >> region[i];
else cin >> p >> region[i];
if (p != -1) adj[p].push_back(i);
}
dfs(1, -1);
FOR(i, 1, numNode) {
regions[region[i]].push_back({in[i], i});
nodes.push_back({i, region[i], in[i]});
}
in[0] = INT_MAX;
FOR(r, 1, numRegions) {
regions[r].push_back({in[0], 0});
sort(regions[r].begin(), regions[r].end());
}
sort(nodes.begin(), nodes.end());
// precomputing for queries that has R1 >= BLOCK - O(BLOCK * N)
int cnt = 0;
FOR(r, 1, numRegions) {
int m = regions[r].size();
if (m >= BLOCK) {
id[r] = ++cnt;
int i = 0, j = 0;
while (i < m && j < numNode) {
if (out[regions[r][i].S] < nodes[j].timer) i++;
if (nodes[j].timer >= in[regions[r][i].S] && nodes[j].timer <= out[regions[r][i].S])
ans[cnt][nodes[j].region]++;
j++;
}
}
}
while (numQuery--) {
int r1, r2;
cin >> r1 >> r2;
if ((int)regions[r1].size() >= BLOCK) cout << ans[id[r1]][r2] << "\n";
else {
solve(r1, r2);
}
}
return 0;
}
