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 edge {
  int c, u, v;
  edge() {}
  void read() { scanf("%d%d%d", &u, &v, &c), u--, v--; }
  bool operator<(const edge &rhs) const { return c < rhs.c; }
} E[3005];
int n, p[3005], cnt[3005], sum[3005], tot;
int find(int u) { return u == p[u] ? u : p[u] = find(p[u]); }
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n - 1; ++i) E[i].read();
  for (int i = 0; i < n; ++i) {
    scanf("%d", sum + i);
    cnt[i] = 1, p[i] = i, tot += sum[i];
  }
  sort(E, E + n - 1);
  int ans = E[0].c;
  for (int i = 0; i < n - 1; ++i) {
    bool ok = 1;
    for (int j = i; j < n - 1; ++j) {
      if (E[j].c != E[i].c) {
        ans = E[j].c;
        i = j - 1;
        break;
      }
      int x = find(E[j].u), y = find(E[j].v);
      cnt[x] += cnt[y];
      sum[x] += sum[y];
      p[y] = x;
      if (cnt[x] > tot - sum[x]) {
        ok = 0;
        break;
      }
    }
    if (!ok) break;
  }
  printf("%d\n", ans);
}