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;
inline int Read() {
  char c = getchar();
  int num = 0;
  while ('0' > c || c > '9') c = getchar();
  while ('0' <= c && c <= '9') num = num * 10 + c - '0', c = getchar();
  return (num);
}
struct Edge {
  int x, y, w;
} e[3010];
int lm[3010], sz[3010], fa[3010], n, ans;
inline int Get(int x) {
  if (fa[x] == x) return (x);
  return (fa[x] = Get(fa[x]));
}
inline bool Check() {
  int nm = 0;
  for (int i = 1; i <= n; i++)
    if (Get(i) == i) nm += min(n - sz[i], lm[i]);
  return (nm >= n);
}
bool Cmpw(Edge a, Edge b) { return (a.w < b.w); }
int main() {
  n = Read();
  for (int i = 1; i < n; i++) e[i].x = Read(), e[i].y = Read(), e[i].w = Read();
  for (int i = 1; i <= n; i++) lm[i] = Read(), sz[i] = 1, fa[i] = i;
  sort(e + 1, e + n, Cmpw);
  for (int i = 1, j = 1; i < n; i = j) {
    for (; e[i].w == e[j].w; j < n) j++;
    for (int k = i; k < j; k++) {
      int fx = Get(e[k].x), fy = Get(e[k].y);
      sz[fx] += sz[fy], fa[fy] = fx, lm[fx] += lm[fy];
    }
    ans = e[i].w;
    if (!Check()) break;
  }
  printf("%d\n", ans);
}