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;
int a[3030], b[3030], c[3030], sz[3030], x[3030];
int root[3030], uu[3030];
int find(int v) {
  if (root[v] == v) return v;
  return root[v] = find(root[v]);
}
int main() {
  int n;
  cin >> n;
  int tot = 0;
  for (int i = 0; i < n - 1; i++) {
    scanf("%d%d%d", &a[i], &b[i], &c[i]);
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", uu + i);
    tot += uu[i];
  }
  int st = 0, ed = 100000;
  while (ed - st > 1) {
    int mid = (ed + st) / 2;
    for (int i = 1; i <= n; i++) root[i] = i, sz[i] = 1, x[i] = uu[i];
    int f = 0;
    for (int i = 0; i < n - 1; i++) {
      if (c[i] >= mid) continue;
      int u = find(a[i]), v = find(b[i]);
      if (u == v) continue;
      sz[u] += sz[v];
      root[v] = u;
      x[u] += x[v];
    }
    for (int i = 1; i <= n; i++) {
      if (find(i) == i and tot - x[i] < sz[i]) f = 1;
    }
    if (f)
      ed = mid;
    else
      st = mid;
  }
  printf("%d", st);
  return 0;
}