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;
int root[3001], cat[3001], val[3001], tot[3001], n;
int c[3001];
vector<pair<int, int> > adj[3001];
inline int find_root(int u) {
  if (u == root[u]) return u;
  return root[u] = find_root(root[u]);
}
void dfs(int u, int par, int limit) {
  pair<int, int> e;
  int i;
  for (i = 0; i < adj[u].size(); i++) {
    e = adj[u][i];
    if (e.first == par) continue;
    if (e.second < limit) root[find_root(u)] = find_root(e.first);
    dfs(e.first, u, limit);
  }
}
bool can(int md) {
  for (int i = 1; i <= n; i++) {
    root[i] = i;
    cat[i] = 0;
    tot[i] = 0;
  }
  dfs(1, 0, md);
  for (int i = 1; i <= n; i++) {
    root[i] = find_root(i);
    cat[root[i]]++;
    tot[root[i]] += c[i];
  }
  int maxL = 1;
  int sum = 0;
  for (int i = 1; i <= n; i++) {
    if (i == root[i]) {
      sum += tot[i];
      if (cat[i] > cat[maxL]) maxL = i;
    }
  }
  return (sum - cat[maxL] >= tot[maxL]);
}
int main() {
  int i, u, v, w;
  scanf("%d", &n);
  for (i = 0; i < n - 1; i++) {
    scanf("%d%d%d", &u, &v, &w);
    adj[u].push_back(make_pair(v, w));
    adj[v].push_back(make_pair(u, w));
    val[i] = w;
  }
  for (int i = 1; i <= n; i++) scanf("%d", c + i);
  sort(val, val + n - 1);
  int ed = unique(val, val + n - 1) - val, st = 0;
  while (ed - st > 1) {
    int md = ed + st >> 1;
    if (can(val[md]))
      st = md;
    else
      ed = md;
  }
  printf("%d\n", val[st]);
  return 0;
}