#include <bits/stdc++.h>
using namespace std;
// {{{
#define rep(i, a, n) for(int i=a; i<n; ++i)
#define per(i, a, n) for(int i=a-1; i>=n; --i)
typedef vector<int> ivec;
typedef long long ll;
typedef pair<int, int> ipair;
typedef vector<vector<ipair> > adjlist;
#define x first
#define y second
#define ITER(A) A.begin(), A.end()
#define REV(A) A.rbegin(), A.rend()
#define pb push_back
#define MAXINT numeric_limits<int>::max()
#define popcount builtin_popcount
#define clz __builtin_clz
#define ctz __builtin_ctz
#define NEW new(nothrow)
// }}}
// AMAX
static inline void amax (int &a, int b) {/*{{{*/
if (a < b) a = b;
}
static inline void amin (int &a, int b) {
if (a > b) a = b;
}/*}}}*/
#define MAXN 110
int n, v[MAXN], t[MAXN], d[MAXN], par[MAXN], dp[MAXN];
adjlist adj;
void dfs (int u, int p) {
par[u] = p;
for (auto it : adj[u]) {
int v = it.x, w = it.y;
if (v != p) {
d[v] = d[u] + w;
dfs(v, u);
}
}
}
vector<ipair> cht;
void dfs_dp (int u, int p) {
dp[u] = t[u] + v[u]*d[u];
if (cht.size() != 0) { /*cht query*/
int l = 0, r = cht.size() - 1;
while (l < r) {
int m = (l+r) / 2;
ipair e1 = cht[m], e2 = cht[m+1];
if ( e1.x * v[u] + e1.y <= e2.x * v[u] + e2.y ) l = m+1;
else r = m;
}
dp[u] -= cht[l].x * v[u] + cht[l].y;
}
stack<ipair> removed;
cht.pb({ d[u], -dp[u] });
while (cht.size() >= 3) {/*cht update*/
int n = cht.size();
ipair a = cht[n-3], b = cht[n-2], c = cht[n-1];
if ( (b.x - c.x) * (b.y - a.y) >= (a.x - b.x) * (c.y - b.y) ) {
cht.pop_back(); cht.pop_back();
cht.pb(c);
removed.push(b);
} else break;
}
for (auto it : adj[u]) if (it.x != p)
dfs_dp(it.x, u);
/*cht restore*/
cht.pop_back();
while (!removed.empty())
cht.pb( removed.top() ), removed.pop();
}
int main () {
scanf("%d", &n);
adj.resize(n+1);
rep (i, 0, n-1) {
int u, v, w; scanf("%d%d%d", &u, &v, &w);
adj[u].pb({ v, w });
adj[v].pb({ u, w });
}
rep(i, 2, n+1) scanf("%d%d", &t[i], &v[i]);
//calculate d[]
d[1] = 0; par[1] = -1;
dfs(1, -1);
//tested
/*
* dp[i] : solution for city i
* dp[1] = 0
* dp[i] = t[i] + v[i]*d[i] - max{j ancestor of k}{ u[i]*d[j] - dp[j] }
*/
// O(N^2){{{
dp[1] = 0;
queue<int> q;
for (auto it : adj[1]) q.push(it.x);
while (!q.empty()) {
int i = q.front(); q.pop();
dp[i] = t[i] + v[i]*d[i];
int temp = (int)-1e9, j = par[i];
while (j != -1) {
amax( temp, v[i]*d[j] - dp[j] );
j = par[j];
}
dp[i] -= temp;
for (auto it : adj[i]) if (it.x != par[i])
q.push(it.x);
}/*}}}*/
rep (i, 2, n+1) printf("%d ", dp[i]);
printf("\n");
// CHT
// dfs_dp(1, -1);
rep (i, 2, n+1) printf("%d ", dp[i]);
return 0;
}
