#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
template<typename T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef long long int ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL)
#define pb push_back
#define endl '\n'
#define sz(a) (int)a.size()
#define setbits(x) __builtin_popcountll(x)
#define ff first
#define ss second
#define conts continue
#define ceil2(x,y) ((x+y-1)/(y))
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define yes cout << "Yes" << endl
#define no cout << "No" << endl
#define rep(i,n) for(int i = 0; i < n; ++i)
#define rep1(i,n) for(int i = 1; i <= n; ++i)
#define rev(i,s,e) for(int i = s; i >= e; --i)
#define trav(i,a) for(auto &i : a)
template<typename T>
void amin(T &a, T b) {
a = min(a,b);
}
template<typename T>
void amax(T &a, T b) {
a = max(a,b);
}
#ifdef LOCAL
#include "debug.h"
#else
#define debug(x) 42
#endif
/*
refs:
https://codeforces.com/blog/entry/88748?#comment-774017 (thread)
got the key idea, but didnt know how to implement in a clean way
implementation idea:
represent the subtree cost @u as a staircase function using a map
can be merged efficiently using small to large
*/
const int MOD = 1e9 + 7;
const int N = 5e3 + 5;
const int inf1 = int(1e9) + 5;
const ll inf2 = ll(1e18) + 5;
vector<ll> adj[N];
vector<pll> a(N);
ll dp[N][N];
ll n;
void dfs(ll u){
trav(v,adj[u]){
dfs(v);
}
auto [h,c] = a[u];
rev(j,n,1){
dp[u][j] = c;
if(j == h) dp[u][j] = 0;
trav(v,adj[u]){
dp[u][j] += dp[v][j];
}
amin(dp[u][j],dp[u][j+1]);
}
}
void solve(int test_case)
{
cin >> n;
rep1(i,n){
ll p,h,c; cin >> p >> h >> c;
if(p != i){
adj[p].pb(i);
}
a[i] = {h,c};
}
vector<ll> b;
rep1(i,n) b.pb(a[i].ff);
sort(all(b));
rep1(i,n) a[i].ff = lower_bound(all(b),a[i].ff)-b.begin()+1;
memset(dp,0x3f,sizeof dp);
dfs(1);
ll ans = inf2;
rep1(j,n) amin(ans,dp[1][j]);
cout << ans << endl;
}
int main()
{
fastio;
int t = 1;
// cin >> t;
rep1(i, t) {
solve(i);
}
return 0;
}
