#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll INF = 1e9+7;
const ll MOD = 998244353;
typedef pair<ll,ll> ii;
#define iii pair<ll,ii>
#define id pair<ll,vector<ll> >
#define f(i,a,b) for(ll i = a;i < b;i++)
#define pb push_back
#define vll vector<ll>
#define F first
#define S second
#define all(x) (x).begin(), (x).end()
///I hope I will get uprating and don't make mistakes
///I will never stop programming
///sqrt(-1) Love C++
///Please don't hack me
///@TheofanisOrfanou Theo830
///Think different approaches (bs,dp,greedy,graphs,shortest paths,mst)
///Stay Calm
///Look for special cases
///Beware of overflow and array bounds
///Think the problem backwards
///Training
#include "tree.h"
int n;
ll leaf = 0;
std::vector<int> p, w;
vector<vector<int> >adj;
ll l,r;
ii solve(ll idx){
ii ans = {0,0};
ll sum = 0;
for(auto x:adj[idx]){
ii res = solve(x);
ans.S += res.S;
sum += res.F;
}
ans.F = sum;
ans.F = min(ans.F,r);
if(sum == 0){
ans = {l,l * w[idx]};
}
else if(sum > r){
ans.S += (sum - r) * w[idx];
}
return ans;
}
void init(std::vector<int> P, std::vector<int> W) {
p = P;
w = W;
n = (int)p.size();
adj.assign(n+5,vector<int>());
f(i,1,n){
adj[p[i]].pb(i);
}
f(i,0,n){
if(adj[i].size() == 0){
leaf++;
}
}
}
long long query(int L, int R){
l = L;
r = R;
return solve(0).S;
}
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