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;
template <typename T>
inline bool chkmax(T &a, T b) {
  return a < b ? a = b, 1 : 0;
}
template <typename T>
inline bool chkmin(T &a, T b) {
  return b < a ? a = b, 1 : 0;
}
template <typename T>
inline T read() {
  static T _, __;
  static char c_;
  for (_ = 0, __ = 1, c_ = getchar(); c_ < '0' || c_ > '9'; c_ = getchar())
    if (c_ == '-') __ = -1;
  for (; c_ >= '0' && c_ <= '9'; c_ = getchar())
    _ = (_ << 1) + (_ << 3) + (c_ ^ 48);
  return _ * __;
}
const int maxn = 3005;
int n, fa[maxn], sz[maxn], in[maxn], sum;
pair<int, pair<int, int> > E[maxn];
int find(int x) { return x == fa[x] ? x : fa[x] = find(fa[x]); }
int main() {
  n = read<int>();
  for (register int i = (1), i_end_ = (n - 1); i <= i_end_; ++i)
    E[i].second.first = read<int>(), E[i].second.second = read<int>(),
    E[i].first = read<int>();
  sort(E + 1, E + n);
  for (register int i = (1), i_end_ = (n); i <= i_end_; ++i)
    fa[i] = i, sz[i] = 1, sum += in[i] = read<int>();
  for (register int i = (1), i_end_ = (n - 1); i <= i_end_; ++i) {
    int x = find(E[i].second.first), y = find(E[i].second.second);
    fa[y] = x, sz[x] += sz[y], in[x] += in[y];
    if (sum - in[x] < sz[x]) return printf("%d\n", E[i].first), 0;
  }
  puts("0");
  return 0;
}