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;
long long read() {
  long long x = 0, f = 0;
  char ch = getchar();
  while (!isdigit(ch)) f = ch == '-', ch = getchar();
  while (isdigit(ch)) x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();
  return f ? -x : x;
}
const int N = 200005;
int n;
int w[N];
vector<pair<int, int> > e[N];
double ans = 1e30, tmp;
int ansid = 0;
int vis[N];
int size[N], mx[N];
int Size, RT;
double dao[N];
void getroot(int x, int pre = 0) {
  size[x] = 1, mx[x] = 0;
  for (auto E : e[x])
    if (E.first != pre && !vis[E.first]) {
      int y = E.first;
      getroot(y, x);
      size[x] += size[y];
      mx[x] = max(mx[x], size[y]);
    }
  mx[x] = max(mx[x], Size - size[x]);
  if (!RT || mx[x] < mx[RT]) RT = x;
}
void dfs(int x, int pre, double d) {
  dao[x] = sqrt(d) * w[x];
  tmp += pow(d, 1.5) * w[x];
  for (auto E : e[x])
    if (E.first != pre) {
      int y = E.first;
      dfs(y, x, d + E.second);
      dao[x] += dao[y];
    }
}
void Solve(int x) {
  RT = 0;
  getroot(x);
  x = RT;
  tmp = 0;
  dfs(x, 0, 0);
  if (tmp < ans) ans = tmp, ansid = x;
  vis[x] = 1;
  for (auto E : e[x]) {
    int y = E.first;
    if (vis[y]) continue;
    if (size[y] > size[x]) size[y] = Size - size[x];
    double d = dao[x] - dao[y] * 2;
    if (d < -1e-6) {
      Size = size[y], Solve(y);
      break;
    }
  }
}
int main() {
  n = read();
  for (int i = (1); i <= (n); i++) w[i] = read();
  for (int i = (1); i <= (n - 1); i++) {
    int x = read(), y = read(), z = read();
    e[x].push_back(make_pair(y, z));
    e[y].push_back(make_pair(x, z));
  }
  Size = n, Solve(1);
  printf("%d %.8lf\n", ansid, ans);
  return 0;
}
