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 std::sort;
using std::vector;
int boss[3000];
int w[3000];
int c[3000];
struct Edge {
  int u, v, w;
} E[3000];
bool ecmp(Edge x, Edge y) { return x.w < y.w; }
int find(int x) {
  if (boss[x] != x) boss[x] = find(boss[x]);
  return boss[x];
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n - 1; ++i) {
    int tu, tv, tw;
    scanf("%d%d%d", &tu, &tv, &tw);
    Edge tmp = (Edge){tu - 1, tv - 1, tw};
    E[i] = tmp;
  }
  sort(E, E + n - 1, ecmp);
  int ans = E[0].w;
  for (int i = 0; i < n; ++i) boss[i] = i;
  int sum = 0;
  for (int i = 0; i < n; ++i) {
    scanf("%d", &w[i]);
    sum += w[i];
    c[i] = 1;
  }
  for (int i = 0; i < n - 1; ++i) {
    int bu = find(E[i].u);
    int bv = find(E[i].v);
    boss[bv] = bu;
    w[bu] += w[bv];
    c[bu] += c[bv];
    if (c[bu] > sum - w[bu])
      break;
    else
      ans = E[i + 1].w;
  }
  printf("%d\n", ans);
  return 0;
}