/*
Can store atmost 1e7 numbers of ll, and abuot 2e7 of int in memory
Fastest way to compute answer for any two region is to use Virtual Tree Trick.
(r1,r2) can be done in O( (sz[r1]+sz[r2]) lg n) , lgn is for sorting,
For computing for all pair, We can do it we O(N*R)
//
*/
#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int N=2e5+1,R=3000+1;
int n,r,q,reg[N],dp[R],ans[R][R];
vector<int> ma[N],sub[N];
void dfs(int x,int p=-1)
{
dp[reg[x]]++;
for(int r1=reg[x],r2=1;r2<=r;r2++) // (N*R)
{
ans[r1][r2]+=dp[r2]-(r1==r2); // how many node such there are parent of x and region is r2
}
for(auto y:ma[x])
{
dfs(y,x);
}
dp[reg[x]]--;
}
int main()
{
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cin>>n>>r>>q;
for(int i=1;i<=n;i++)
{
if(i==1)
{
cin>>reg[i];
continue;
}
int x;
cin>>x>>reg[i];
ma[x].push_back(i);
}
for(int i=1;i<=n;i++)
{
sub[reg[i]].push_back(i);
}
// r<=500
// thus we can do child[node][region]
dfs(1);
// get summation sz[r1]+sz[r2] (r1,r2) summation(sz[r])*R // N*R
// R*R ans[r1][r2] = sum(dp[x][r1]) x in sub[r1] sz[r1]
//
// for(int r1=1;r1<=r;r1++)
// {
// for(auto x:sub[r1])
// {
// for(int r2=1;r2<=r;r2++)
// {
// ans[r1][r2]+=dp[x][r2]-(r1==r2);
// }
// }
// }
while(q--)
{
int r1,r2;
cin>>r1>>r2;
cout<<ans[r2][r1]<<endl;
}
}
