Problem: DZY loves planting, and he enjoys solving tree problems.

DZY has a weighted tree (connected undirected graph without cycles) containing n nodes (they are numbered from 1 to n). He defines the function g(x, y) (1 ≤ x, y ≤ n) as the longest edge in the shortest path between nodes x and y. Specially g(z, z) = 0 for every z.

For every integer sequence p1, p2, ..., pn (1 ≤ pi ≤ n), DZY defines f(p) as <image>. 

DZY wants to find such a sequence p that f(p) has maximum possible value. But there is one more restriction: the element j can appear in p at most xj times.

Please, find the maximum possible f(p) under the described restrictions.

Input

The first line contains an integer n (1 ≤ n ≤ 3000).

Each of the next n - 1 lines contains three integers ai, bi, ci (1 ≤ ai, bi ≤ n; 1 ≤ ci ≤ 10000), denoting an edge between ai and bi with length ci. It is guaranteed that these edges form a tree.

Each of the next n lines describes an element of sequence x. The j-th line contains an integer xj (1 ≤ xj ≤ n).

Output

Print a single integer representing the answer.

Examples

Input

4
1 2 1
2 3 2
3 4 3
1
1
1
1


Output

2


Input

4
1 2 1
2 3 2
3 4 3
4
4
4
4


Output

3

Note

In the first sample, one of the optimal p is [4, 3, 2, 1].

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T, typename U>
inline void smin(T &a, U b) {
  if (a > b) a = b;
}
template <typename T, typename U>
inline void smax(T &a, U b) {
  if (a < b) a = b;
}
inline void getint(int &first) {
  char c;
  while (c = getchar(), c > '9' || c < '0')
    ;
  for (first = 0; c >= '0' && c <= '9'; c = getchar())
    first = (first << 1) + (first << 3) + c - '0';
}
inline void print(int first) {
  if (first < 10) {
    putchar('0' + first);
    return;
  }
  print(first / 10);
  putchar(first % 10 + '0');
}
int power(int a, int b, int m, int ans = 1) {
  for (; b; b >>= 1, a = 1LL * a * a % m)
    if (b & 1) ans = 1LL * ans * a % m;
  return ans;
}
pair<pair<int, int>, int> edge[3333];
int root[3333], cnt[3333], ss[3333], a[3333];
void init(int n) {
  for (int i = 1; i <= n; i++) root[i] = i, cnt[i] = 1, ss[i] = a[i];
}
int find_root(int u) { return u == root[u] ? u : root[u] = find_root(root[u]); }
void join(int u, int v) {
  u = find_root(u);
  v = find_root(v);
  if (u == v) return;
  if (cnt[u] < cnt[v]) swap(u, v);
  root[v] = u;
  cnt[u] += cnt[v];
  ss[u] += ss[v];
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    scanf("%d %d %d", &edge[i].first.first, &edge[i].first.second,
          &edge[i].second);
  }
  int sum = 0;
  for (int i = 1; i <= n; i++) scanf("%d", a + i), sum += a[i];
  int st = 0, ed = 10001, mid;
  while (ed - st > 1) {
    mid = st + ed >> 1;
    init(n);
    for (int i = 0; i < n - 1; i++) {
      if (edge[i].second < mid) join(edge[i].first.first, edge[i].first.second);
    }
    int ok = 1;
    for (int i = 1; i <= n && ok; i++)
      if (find_root(i) == i) {
        if (cnt[i] > sum - ss[i]) ok = 0;
      }
    if (ok)
      st = mid;
    else
      ed = mid;
  }
  cout << st << endl;
  return 0;
}