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 void read(int &x) {
  x = 0;
  int p = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') p = -p;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    x = (x << 1) + (x << 3) + (ch - '0'), ch = getchar();
  x = x * p;
}
const int maxn = 200005;
int n, w[maxn], head[maxn], cnt, siz[maxn], vis[maxn], son[maxn], root, maxson;
int ansu;
double res, dao[maxn], totdao, nowres;
struct edge {
  int to, next, v;
} e[maxn << 1];
void add(int x, int y, int z) {
  e[cnt].to = y;
  e[cnt].v = z;
  e[cnt].next = head[x];
  head[x] = cnt++;
}
void getsiz(int u, int pre) {
  siz[u] = 1;
  for (int i = head[u]; i; i = e[i].next) {
    if (e[i].to == pre || vis[e[i].to]) continue;
    getsiz(e[i].to, u);
    siz[u] += siz[e[i].to];
  }
}
void getson(int u, int pre) {
  son[u] = 0;
  for (int i = head[u]; i; i = e[i].next) {
    if (e[i].to == pre || vis[e[i].to]) continue;
    getson(e[i].to, u);
    son[u] = max(son[u], siz[e[i].to]);
  }
  son[u] = max(son[u], maxson - siz[u]);
  if (son[root] > son[u]) root = u;
}
void getdao(int u, int pre, long long d) {
  dao[u] = 0.0;
  for (int i = head[u]; i; i = e[i].next) {
    if (e[i].to == pre) continue;
    getdao(e[i].to, u, d + e[i].v);
    dao[u] += dao[e[i].to];
  }
  dao[u] += 1.5 * w[u] * sqrt(d);
  nowres += 1.0 * w[u] * pow(d, 1.5);
}
void solve(int u) {
  getsiz(u, 0);
  root = 0;
  maxson = siz[u];
  getson(u, 0);
  u = root;
  vis[u] = 1;
  totdao = 0.0;
  nowres = 0.0;
  for (int i = head[u]; i; i = e[i].next) {
    getdao(e[i].to, u, e[i].v);
    totdao += dao[e[i].to];
  }
  if (nowres < res) res = nowres, ansu = u;
  for (int i = head[u]; i; i = e[i].next) {
    if (vis[e[i].to]) continue;
    double x = totdao - 2.0 * dao[e[i].to];
    if (x < 0) {
      solve(e[i].to);
      return;
    }
  }
}
int main() {
  ansu = 0;
  res = 1e20;
  son[0] = 1061109567;
  read(n);
  cnt = 1;
  for (int i = 1; i <= n; ++i) read(w[i]);
  for (int i = 1; i < n; ++i) {
    int x, y, z;
    read(x);
    read(y);
    read(z);
    add(x, y, z);
    add(y, x, z);
  }
  solve(1);
  printf("%d %.10f\n", ansu, res);
}
