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>
int n;
int x[4000];
int adj[4000];
int aim[8000], next[8000], weight[8000];
int visit[4000];
int sumx[4000];
int count[4000];
int ans;
void dfs(int r, int w, int s) {
  visit[r] = 1;
  sumx[s] += x[r];
  count[s] += 1;
  int i = adj[r];
  while (i) {
    if (weight[i] < w && !visit[aim[i]]) dfs(aim[i], w, s);
    i = next[i];
  }
}
int check(int w) {
  int flex = 0;
  for (int i = 0; i < n; ++i) {
    visit[i] = 0;
    count[i] = 0;
    sumx[i] = 0;
  }
  for (int i = 0; i < n; ++i)
    if (!visit[i]) dfs(i, w, i);
  for (int i = 0; i < n; ++i) flex += sumx[i] - count[i];
  for (int i = 0; i < n; ++i)
    if (count[i] > n / 2) {
      if (count[i] - flex + sumx[i] - count[i] >
          (n - flex + sumx[i] - count[i]) / 2)
        return 0;
      break;
    }
  return 1;
}
int main() {
  scanf("%d", &n);
  int a, b, c;
  for (int i = 0; i < n - 1; ++i) {
    scanf("%d %d %d", &a, &b, &c);
    --a;
    --b;
    aim[i * 2 + 1] = b;
    next[i * 2 + 1] = adj[a];
    adj[a] = i * 2 + 1;
    aim[i * 2 + 2] = a;
    next[i * 2 + 2] = adj[b];
    adj[b] = i * 2 + 2;
    weight[i * 2 + 1] = weight[i * 2 + 2] = c;
  }
  for (int i = 0; i < n; ++i) scanf("%d", &x[i]);
  int l = 0, r = 10000;
  int mid;
  while (r - l > 1) {
    mid = (l + r) / 2;
    if (check(mid))
      l = mid;
    else
      r = mid;
  }
  if (check(r))
    ans = r;
  else
    ans = l;
  printf("%d\n", ans);
  return 0;
}