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;
inline int getch() {
  static const int len = 1 << 20;
  static char buf[len], *st, *ed;
  if (st == ed) ed = buf + fread(st = buf, 1, len, stdin);
  return st == ed ? EOF : *st++;
}
template <typename T>
inline void read(T& x) {
  static int c;
  x = 0, c = getch();
  while (isspace(c)) c = getch();
  while (isdigit(c)) x = x * 10 + c - '0', c = getch();
  return;
}
template <typename T, typename... Args>
void read(T& x, Args&... args) {
  return read(x), read(args...);
}
const int maxn = 3003;
tuple<int, int, int> edge[maxn];
int n, m, u, v, w, f[maxn], s[maxn], x[maxn], cur, sum, ans;
int find(int x) { return f[x] == x ? x : f[x] = find(f[x]); }
int merge(int p, int q) {
  p = find(p), q = find(q);
  if (s[p] > s[q])
    return s[p] += s[q], x[p] += x[q], f[q] = p;
  else
    return s[q] += s[p], x[q] += x[p], f[p] = q;
}
int main() {
  read(n);
  for (int i = 1; i < n; i++) read(u, v, w), edge[i] = tie(w, u, v);
  for (int i = 1; i <= n; i++) f[i] = i, s[i] = 1, read(x[i]), sum += x[i];
  sort(edge + 1, edge + n);
  for (int i = 1; i < n; i++) {
    tie(w, u, v) = edge[i], cur = merge(u, v);
    if (s[cur] > sum - x[cur]) return printf("%d\n", w), 0;
  }
  return printf("%d\n", w), 0;
}