#include <bits/stdc++.h>
#define FOR(i, x, y) for (int i = x; i < y; i++)
typedef long long ll;
using namespace std;
vector<pair<int, ll>> graph[1000001], stck, cycle;
ll discarded, ans = 0, cmp_ans, dp[1000001], p[4][1000001];
bool processed[1000001], on_stck[1000001], on_cycle[1000001];
bool find_cycle(int node, pair<int, ll> parent = {0, 0}) {
on_stck[node] = true;
for (pair<int, ll> i : graph[node]) if (i != parent) {
stck.push_back({node, i.second});
if (on_stck[i.first]) {
cycle.push_back({i.first, 0});
on_cycle[i.first] = true;
while (stck.back().first != i.first) {
cycle.push_back(stck.back());
on_cycle[stck.back().first] = true;
stck.pop_back();
}
discarded = stck.back().second;
return true;
}
if (find_cycle(i.first, {node, i.second})) return true;
stck.pop_back();
on_stck[i.first] = false;
}
return false;
}
void calc_tree(int node, int parent = 0) {
processed[node] = true;
ll mx = 0, sc = 0;
for (pair<int, ll> i : graph[node]) if (i.first != parent && !on_cycle[i.first]) {
calc_tree(i.first, node);
if (dp[i.first] + i.second > mx) sc = mx, mx = dp[i.first] + i.second;
else if (dp[i.first] + i.second > sc) sc = dp[i.first] + i.second;
}
dp[node] = mx;
cmp_ans = max(cmp_ans, mx + sc);
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
int n;
cin >> n;
FOR(i, 1, n + 1) {
int v;
ll l;
cin >> v >> l;
graph[i].push_back({v, l});
graph[v].push_back({i, l});
}
FOR(i, 1, n + 1) if (!processed[i]) {
stck.clear(); cycle.clear();
find_cycle(i);
cmp_ans = 0;
int m = cycle.size();
FOR(j, 1, m) p[0][j] = p[0][j - 1] + cycle[j].second;
for (int j = m - 2; ~j; j--) p[1][j] = p[1][j + 1] + cycle[j + 1].second;
FOR(j, 0, m) {
calc_tree(cycle[j].first);
p[2][j] = dp[cycle[j].first] - p[0][j];
p[3][j] = dp[cycle[j].first] + p[0][j];
p[4][j] = dp[cycle[j].first] + p[1][j];
}
ll mx = p[2][0];
FOR(j, 1, m) {
cmp_ans = max(cmp_ans, p[3][j] + mx);
mx = max(mx, p[2][j]);
}
mx = p[4][m - 1];
for (int j = m - 2; ~j; j--) {
cmp_ans = max(cmp_ans, p[3][j] + mx + discarded);
mx = max(mx, p[4][j]);
}
ans += cmp_ans;
FOR(j, 0, 5) FOR(k, 0, m) p[j][k] = 0;
}
cout << ans;
return 0;
}
