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;
struct edg {
  int s, e, v;
  bool operator<(const edg& c) const { return v < c.v; }
} ba[3333];
int n, xi[3333], par[3333], cnt[3333];
int root(int ix) {
  while (par[ix]) ix = par[ix];
  return ix;
}
int main() {
  int i, j, k, sum;
  scanf("%d", &n);
  for (i = 0; i < n - 1; i++) scanf("%d%d%d", &ba[i].s, &ba[i].e, &ba[i].v);
  sort(ba, ba + n - 1);
  for (i = 1; i <= n; i++) scanf("%d", &xi[i]), cnt[i] = 1;
  for (i = 0; i < n - 1; i++) {
    for (j = 1; j <= n; j++) {
      if (!par[j] && cnt[j] > n / 2) break;
    }
    if (j <= n) {
      sum = 0;
      for (k = 1; k <= n; k++) {
        if (root(k) != j) sum += xi[k];
      }
      if (sum < cnt[j]) break;
    }
    j = root(ba[i].s), k = root(ba[i].e);
    if (cnt[j] < cnt[k])
      par[j] = k, cnt[k] += cnt[j];
    else
      par[k] = j, cnt[j] += cnt[k];
  }
  printf("%d", ba[i - 1].v);
  return 0;
}