3
1 1 5
1 3 42
1 2 42
 2
42 23
1 2 10
 10
45792697 27183921 4791060 77070594 27286110 66957065 19283638 2944837 26775511 65379701
5 8 25
8 10 94
1 4 14
9 7 96
4 2 16
2 6 58
2 3 99
6 8 21
8 9 87
 5
3 1 2 6 5
1 2 3
2 3 1
4 3 9
5 3 1
 10
26 31 21 57 56 44 54 58 79 97
2 1 277
1 8 91
1 7 380
4 1 649
10 5 799
10 1 940
8 6 627
6 3 177
9 6 685
 10
8 100 62 38 26 51 26 23 77 32
6 8 29
4 1 82
7 8 66
7 10 67
8 9 98
6 1 14
1 5 48
3 1 58
2 9 90
 2
23 42
1 2 10
 1
42
 3
1 5 1
1 3 42
1 2 42
 10
1 1 1 1 1 1 1 1 1 1
1 2 5
1 3 5
1 4 5
1 5 5
1 6 5
1 7 5
1 8 5
1 9 5
1 10 5
 3
1 1 1
1 2 42
2 3 42
 3
5 1 1
1 3 42
1 2 24

3 1042.0638260857 1 727.3238618387 6 283959805416.40496826
 3 192.0000000000 1 13952642.4776933547 8 544327.1927847408 2 727.3238618387 1 0.0000000000 2 1042.0638260857 1 100.6230589875 2 544.3822186663 1 389.7666169867
#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 1) + (x << 3) + (c ^ 48);
    c = getchar();
  }
  return f == 1 ? x : -x;
}
const int N = 2e5 + 4;
vector<pair<int, int> > e[N];
int n, rt, ans1, a[N], vis[N], siz[N], son[N];
double sd, sum, ans2, d[N];
int findsz(int x, int fa) {
  int sz = 1;
  for (auto i : e[x]) {
    int v = i.first;
    if (vis[v] || v == fa) continue;
    sz += findsz(v, x);
  }
  return sz;
}
void findrt(int x, int fa, int S) {
  siz[x] = 1;
  son[x] = 0;
  for (auto i : e[x]) {
    int v = i.first;
    if (vis[v] || v == fa) continue;
    findrt(v, x, S);
    siz[x] += siz[v];
    son[x] = max(son[x], siz[v]);
  }
  son[x] = max(son[x], S - siz[x]);
  if (!rt || vis[rt] || son[x] < son[rt]) rt = x;
}
void calc(int x, int fa, int o, int z) {
  sum += pow(z, 1.5) * a[x];
  d[o] += sqrt(z) * 1.5 * a[x];
  for (auto i : e[x]) {
    int v = i.first;
    if (v == fa) continue;
    calc(v, x, o, z + i.second);
  }
}
void solve(int x) {
  findrt(x, 0, findsz(x, 0));
  if (vis[rt] || !rt) return;
  x = rt;
  vis[x] = 1;
  sum = sd = 0;
  for (auto i : e[x]) {
    int v = i.first;
    d[v] = 0;
    calc(v, x, v, i.second);
    sd += d[v];
  }
  if (sum < ans2) {
    ans1 = x;
    ans2 = sum;
  }
  for (auto i : e[x]) {
    int v = i.first;
    if (sd - d[v] * 2 >= 0) continue;
    solve(v);
  }
}
int main() {
  n = read();
  for (int i = 1; i <= n; i++) a[i] = read();
  for (int i = 1, u, v, w; i < n; i++) {
    u = read();
    v = read();
    w = read();
    e[u].push_back(make_pair(v, w));
    e[v].push_back(make_pair(u, w));
  }
  ans2 = 1e20;
  solve(1);
  printf("%d %.8lf", ans1, ans2);
  return (0 - 0);
}
