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;
const int MAXN = 5000;
pair<int, pair<int, int> > lis[MAXN];
int n;
int pre[MAXN];
int p[MAXN], siz[MAXN];
int findPre(int p) {
  if (pre[p] != pre[pre[p]]) {
    pre[p] = findPre(pre[p]);
  }
  return pre[p];
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n - 1; ++i) {
    scanf("%d%d%d", &lis[i].second.first, &lis[i].second.second, &lis[i].first);
    lis[i].second.first--;
    lis[i].second.second--;
  }
  int tot = 0;
  for (int i = 0; i < n; ++i) {
    scanf("%d", &p[i]);
    tot += p[i];
    pre[i] = i;
    siz[i] = 1;
  }
  if (n == 1) {
    puts("0");
  } else {
    sort(lis, lis + (n - 1));
    for (int i = 0; i < n - 1; ++i) {
      int a = findPre(lis[i].second.first);
      int b = findPre(lis[i].second.second);
      pre[b] = a;
      siz[a] += siz[b];
      p[a] += p[b];
      if (siz[a] > (tot - p[a])) {
        printf("%d\n", lis[i].first);
        return 0;
      }
    }
  }
  return 0;
}