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>
const int MAXN = 3010;
struct _ {
  int b, e, v;
  bool operator<(const _& b) const { return v < b.v; }
} es[MAXN];
int fa[MAXN], sz[MAXN], xs[MAXN], n;
int find(int x) { return x == fa[x] ? x : fa[x] = find(fa[x]); }
int main() {
  std::cin >> n;
  for (int i = 1; i < n; ++i) std::cin >> es[i].b >> es[i].e >> es[i].v;
  std::sort(es + 1, es + n);
  int tox = 0;
  for (int i = 1; i <= n; ++i)
    std::cin >> xs[i], sz[fa[i] = i] = 1, tox += xs[i];
  int ans = es[1].v;
  for (int i = 1; i < n; ++i) {
    int x = find(es[i].b), y = find(es[i].e);
    sz[y] += sz[x], xs[y] += xs[x];
    fa[x] = y;
    if ((tox - xs[y]) < sz[y]) break;
    ans = es[i + 1].v;
  }
  std::cout << ans << std::endl;
  return 0;
}