/*input
6 3 4
1
1 2
1 3
2 3
2 3
5 1
1 2
1 3
2 3
3 1
*/
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
namespace my_template {
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;
typedef pair<int, int> pi;
typedef pair<ll, ll> pl;
typedef pair<ld, ld> pd;
typedef vector<int> vi;
typedef vector<vi> vii;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<vl> vll;
typedef vector<pi> vpi;
typedef vector<vpi> vpii;
typedef vector<pl> vpl;
typedef vector<cd> vcd;
typedef vector<pd> vpd;
typedef vector<bool> vb;
typedef vector<vb> vbb;
typedef std::string str;
typedef std::vector<str> vs;
#define x first
#define y second
#define debug(...) cout<<"["<<#__VA_ARGS__<<": "<<__VA_ARGS__<<"]\n"
const ld PI = 3.14159265358979323846264338327950288419716939937510582097494L;
template<typename T>
pair<T, T> operator+(const pair<T, T> &a, const pair<T, T> &b) { return pair<T, T>(a.x + b.x, a.y + b.y); }
template<typename T>
pair<T, T> operator-(const pair<T, T> &a, const pair<T, T> &b) { return pair<T, T>(a.x - b.x, a.y - b.y); }
template<typename T>
T operator*(const pair<T, T> &a, const pair<T, T> &b) { return (a.x * b.x + a.y * b.y); }
template<typename T>
T operator^(const pair<T, T> &a, const pair<T, T> &b) { return (a.x * b.y - a.y * b.x); }
template<typename T>
void print(vector<T> vec, string name = "") {
cout << name;
for (auto u : vec)
cout << u << ' ';
cout << '\n';
}
}
using namespace my_template;
const int MOD = 1000000007;
const ll INF = std::numeric_limits<ll>::max();
const int MX = 200101;
int N, R, Q;
vi edges[MX];
vi kas;
vii kokie;
vi timeIn;
vi timeOut;
vi atgal;
vi sz;
vii ats(501, vi(501, 0));
int piv = 0;
int dfs(int x) {
atgal[piv] = x;
timeIn[x] = piv++;
sz[x] = 1;
for (auto u : edges[x])
sz[x] += dfs(u);
timeOut[x] = piv;
return sz[x];
}
bool anc(int a, int b) { // a is ancestor of b
return timeIn[a] <= timeIn[b] && timeOut[b] <= timeOut[a];
}
void dfs(int x, vi &curr) {
for (auto u : edges[x]) {
vi naujas(R + 1, 0);
dfs(u, naujas);
for (int i = 1; i <= R; ++i)
curr[i] += naujas[i];
}
// have curr
for (int i = 1; i <= R; ++i)
ats[kas[x]][i] += curr[i];
curr[kas[x]]++;
}
void precompute() {
vi pivas(R + 1, 0);
dfs(1, pivas);
}
int main() {
ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
cin >> N >> R >> Q;
kas.resize(N + 1);
kokie.resize(R + 1);
timeIn.resize(N + 1);
timeOut.resize(N + 1);
sz.resize(N + 1);
atgal.resize(N + 1);
cin >> kas[1];
kokie[kas[1]].emplace_back(1);
for (int i = 2; i <= N; ++i)
{
int a; cin >> a >> kas[i];
edges[a].push_back(i);
kokie[kas[i]].emplace_back(i);
}
dfs(1);
if (R > 500) {
for (int i = 0; i < Q; ++i)
{
int r1, r2;
cin >> r1 >> r2;
int ans = 0;
for (auto a : kokie[r1]) {
for (auto b : kokie[r2]) {
ans += anc(a, b);
}
}
cout << ans << endl;
}
return 0;
} else {
assert(false);
}
// R <= 500
precompute();
for (int i = 0; i < Q; ++i)
{
int r1, r2;
cin >> r1 >> r2;
cout << ats[r1][r2] << endl;
}
}
/* Look for:
* special cases (n=1?)
* overflow (ll vs int?)
* the exact constraints (multiple sets are too slow for n=10^6 :( )
* array bounds
*/
