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 N = 3005;
int x[N], y[N], z[N], a[N], s, size1[N], size2[N], n, f[N];
int find(int x) {
  if (x == f[x]) return x;
  return f[x] = find(f[x]);
}
int pd(int val) {
  for (int i = 1; i <= n; i++) f[i] = i;
  for (int i = 1; i < n; i++)
    if (z[i] < val) f[find(x[i])] = find(y[i]);
  memset(size1, 0, sizeof size1);
  memset(size2, 0, sizeof size2);
  for (int i = 1; i <= n; i++) size1[find(i)]++, size2[find(i)] += a[i];
  for (int i = 1; i <= n; i++)
    if (find(i) == i && s - size2[i] < size1[i]) return 0;
  return 1;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) scanf("%d%d%d", &x[i], &y[i], &z[i]);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), s += a[i];
  int l = 0, r = 10000;
  while (l < r) {
    int mid = (l + r + 1) / 2;
    if (pd(mid))
      l = mid;
    else
      r = mid - 1;
  }
  printf("%d\n", l);
}