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 n, ans, total;
int app[3005], ata[3005], sz[3005];
pair<int, pair<int, int> > edge[3005];
int bul(int node) {
  if (ata[node] == node) return node;
  return ata[node] = bul(ata[node]);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++)
    scanf("%d %d %d", &edge[i].second.first, &edge[i].second.second,
          &edge[i].first);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &app[i]);
    total += app[i];
    ata[i] = i;
    sz[i] = 1;
  }
  sort(edge + 1, edge + n);
  for (int i = 1; i < n; i++) {
    int a = bul(edge[i].second.first);
    int b = bul(edge[i].second.second);
    ata[a] = b;
    sz[b] += sz[a];
    app[b] += app[a];
    if (sz[b] > total - app[b]) {
      ans = edge[i].first;
      break;
    }
  }
  printf("%d", ans);
}