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;
const long long mod = 1e9 + 7;
const long long inf = 4e18;
const long double pai = acos(-1);
int n, all, bad;
pair<int, pair<int, int> > e[3009];
int p[100090], need[100009], give[100009];
int get(int node) {
  if (p[node] == node) return node;
  return p[node] = get(p[node]);
}
void merge(int a, int b) {
  a = get(a), b = get(b);
  if (a == b) return;
  p[b] = a;
  need[a] += need[b];
  give[a] += give[b];
  if (need[a] > all - give[a]) bad = 1;
}
int main() {
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int a, b, c;
    cin >> a >> b >> c;
    a--, b--;
    e[i] = {c, {a, b}};
  }
  sort(e, e + n - 1);
  int ans = 0;
  for (int i = 0; i < n; i++) {
    cin >> give[i];
    all += give[i];
    p[i] = i;
    need[i] = 1;
  }
  for (int i = 0; i < n; i++) {
    if (need[i] > all - give[i]) bad = 1;
  }
  if (!bad && n > 1) ans = e[0].first;
  for (int i = 0; i < n - 1; i++) {
    int j = i;
    while (j < n - 1 && e[i].first == e[j].first) {
      merge(e[j].second.first, e[j].second.second);
      j++;
    }
    if (!bad && j < n) ans = e[j].first;
    i = j - 1;
  }
  cout << ans << endl;
}