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 read() {
  register int x = 0;
  register char f = 1, ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') f = 0;
  for (; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + (ch ^ '0');
  return f ? x : -x;
}
int n, fa[100005], sz[100005], ans;
long long a[100005], sum;
struct node {
  int u, v, w;
  bool operator<(const node &res) const { return w < res.w; }
} e[100005];
int find(int x) { return fa[x] == x ? x : (fa[x] = find(fa[x])); }
int merge(int x, int y) {
  int fx = find(x), fy = find(y);
  fa[fy] = fx, sz[fx] += sz[fy], a[fx] += a[fy];
  return sz[fx] <= sum - a[fx];
}
int main() {
  n = read();
  for (register int i = 1; i < n; ++i) e[i] = (node){read(), read(), read()};
  for (register int i = 1; i <= n; ++i)
    sum += a[i] = read(), fa[i] = i, sz[i] = 1;
  std ::sort(e + 1, e + n);
  for (register int i = 1; i < n; ++i)
    if (ans = e[i].w, !merge(e[i].u, e[i].v)) break;
  printf("%d", ans);
}