#include <bits/stdc++.h>
using namespace std;
using pii = pair<int, int>;
#define all(x) (x).begin(), (x).end()
vector<int> values;
vector<int> days;
vector<vector<int>> adj;
int dpf(int node, int limd){
int res = (days[node]<=limd? values[node] : 0);
int res2= 0;
int juice=values[node];
int day=days[node];
for(auto thing:adj[node]){
res+=dpf(thing, min(days[node], limd));
res2+=dpf(thing, min((int)1e9, limd));
}
return max(res, res2);
}
void solve(){
int n, m, k;cin>>n>>m>>k;
values.assign(n+1, 0);
days.assign(n+1, 0);
adj.resize(n+1);
for(int i = 2;i<=n;i++){
int p;cin>>p;
adj[p].push_back(i);
}
for(int i = 0;i<m;i++){
int v, d, w;cin>>v>>d>>w;
values[v]=w;
days[v]=d;
}
cout << dpf(1, (int)1e9) << '\n';
//with sets its total good time merging, but I also need to mantain
//like a double log n full I do have it but its so shitty i mean
//39 minutes got all ideas exceot one
//pretty sure 83 very doable because OOH WAIT
//so returns a set<int>
//when small to large merging then I can also detect needed day
//all values of v distinct thus no vertex >1 fruit that's good
//so for each subtree value, days not important UNLESS two on same day then if I want to cut
//so cutting something annulates everything cut strictly after
//so I will redo dfs, check values after
//if values after contain bad bad bad >me juice, do nothing else cut. so
//dpf(node) should return maximum juice value I can get after day x
//so I should start at the root
//then for each fruit I see (basically at every node)
//I do:SMALL TO LARGE MERGING
//like I will have times
// times are very not good
//so because those are sets I will swap em and redo
//if done correctly with references then a good time
}
signed main(){
ios_base::sync_with_stdio(false);cin.tie(0);
int t = 1;
while (t--)
{
solve();
}
}
