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 n;
pair<int, pair<int, int> > q[3333];
int p[3333], sz[3333], kol[3333];
int maxsz, sum = 0, bad = 0;
int get(int x) {
  if (p[x] == x) return x;
  return p[x] = get(p[x]);
}
void unite(int x, int y) {
  x = get(x);
  y = get(y);
  maxsz = max(maxsz, sz[x] + sz[y]);
  sz[x] += sz[y];
  kol[x] += kol[y];
  if (kol[x] > sum - sz[x]) bad = 1;
  p[y] = x;
}
int main() {
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    scanf("%d%d%d", &q[i].second.first, &q[i].second.second, &q[i].first);
    q[i].second.first--;
    q[i].second.second--;
  }
  sort(q, q + n - 1);
  int ans = 0;
  bad = 0;
  for (int i = 0; i < n; i++) {
    p[i] = i;
    scanf("%d", &sz[i]);
    kol[i] = 1;
    sum += sz[i];
  }
  for (int i = 0; i < n; i++)
    if (kol[i] > sum - sz[i]) bad = 1;
  if (bad == 0 && n > 1) ans = q[0].first;
  for (int i = 0; i < n - 1;) {
    int j = i;
    while (j < n && q[j].first == q[i].first) {
      unite(q[j].second.first, q[j].second.second);
      j++;
    }
    i = j;
    if (bad == 0 && i < n) ans = q[i].first;
    if (bad) break;
  }
  cout << ans << endl;
  return 0;
}