/**
* Solution by Charles Ran (polarity.sh)
* Date: 2025-03-01
* Contest: APIO 2014
* Problem: beads
**/
#include <bits/stdc++.h>
using namespace std;
using ull = unsigned long long;
using ll = long long;
using vi = vector<int>;
using vl = vector<ll>;
using pii = pair<int, int>;
#define pb push_back
#define rep(i, a, b) for(int i = (a); i < (b); ++i)
#define all(x) begin(x), end(x)
#define sz(x) (int)(x).size()
const int MAX_N = 1e5 + 1;
const ll MOD = 1e9 + 7;
void solve(){
int n; cin >> n;
vector<vector<pii>> adj(n);
rep(i, 0, n - 1){
int a, b, c; cin >> a >> b >> c;
--a; --b;
adj[a].pb({b, c});
adj[b].pb({a, c});
}
// { any, cannot abb }
vector<pii> dp(n);
vi par(n);
function<void(int, int, int)> dfs1;
function<void(int)> dfs2;
dfs1 = [&](int node, int parent, int last){
int a = 0;
int mx = INT_MIN;
par[node] = parent;
for (auto &[child, cost] : adj[node]){
if (child == parent){
continue;
}
dfs1(child, node, cost);
a += dp[child].first;
mx = max(mx, cost + dp[child].second - dp[child].first);
}
if (last != -1 && mx != INT_MIN){
mx += a;
mx += last;
} else {
mx = 0;
}
dp[node] = {max(a, mx), a};
};
// when 0 is the starting bead
dfs1(0, -1, -1);
vector<int> brcost(n, 0);
vector<multiset<int>> br(n);
// precompute
rep(i, 0, n){
for (auto &[child, cost] : adj[i]){
if (child == par[i]){
continue;
}
br[i].insert(cost + dp[child].second - dp[child].first);
brcost[i] += dp[child].first;
}
}
int ans = dp[0].second;
// transition to check when each node is the starting bead
dfs2 = [&](int node){
for (auto &[child, cost] : adj[node]){
if (child == par[node]){
continue;
}
// no connect
int x = brcost[node] - dp[child].first;
ans = max(ans, x + dp[child].second);
// connect
int adjustment = cost + dp[child].second - dp[child].first;
// assert(br[node].contains(adjustment));
br[node].erase(br[node].find(adjustment));
int c = *prev(br[node].end());
ans = max(ans, c + x + cost + dp[child].second);
br[node].insert(adjustment);
int a = max(x, c + x + cost);
br[child].insert(cost + x - a);
brcost[child] += a;
dfs2(child);
}
};
dfs2(0);
cout << ans << endl;
// previous direct dp attempt, does not work but maybe possible
// dfs = [&](ll node, ll parent, ll last){
// ll a = INT_MIN, b1 = 0, b2 = 0, c = INT_MIN;
// ll mx1 = INT_MIN, mx2 = INT_MIN;
// ll x = 0;
// for (auto &[child, cost] : adj[node]){
// if (child == parent){
// continue;
// }
// dfs(child, node, cost);
// c = max(c, dp[child][0] - dp[child][2]);
// x += dp[child][2];
// if (last != -1){
// b1 = max(b1, last + cost + dp[child][3] - dp[child][2]);
// b2 = max(b2, last + cost + dp[child][4] - dp[child][2]);
// }
// ll change1 = cost + dp[child][3] - dp[child][2];
// ll change2 = cost + dp[child][4] - dp[child][2];
// a = max(a, change1 + mx2);
// a = max(a, change2 + mx1);
// mx1 = max(mx1, change1);
// mx2 = max(mx2, change2);
// }
// if ((parent == -1 && adj[node].size() == 1) || adj[node].size() > 1){
// c += x;
// } else {
// c = 0;
// }
// if ((parent == -1 && adj[node].size() == 2) || adj[node].size() > 2){
// a += x;
// } else {
// a = 0;
// }
// if (last != -1){
// b1 += x;
// b2 += x;
// } else {
// b1 = 0;
// b2 = 0;
// }
// ll b = max(b1, b2);
// ll ac = max(a, c);
// dp[node] = {max(ac, b), max(ac, b1), max(ac, b2), ac, c};
// };
}
int main(){
ios_base::sync_with_stdio(0);
cin.tie(0); cout.tie(0);
solve();
return 0;
}
