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 tot, N, t[3009], mx[3009], L[3009];
bool ok = 1;
pair<int, pair<int, int> > edge[3009];
int tata(int x) {
  if (t[x] == x) return x;
  return t[x] = tata(t[x]);
}
void unite(int x, int y) {
  x = tata(x), y = tata(y);
  L[x] += L[y];
  mx[x] += mx[y];
  if (tot - mx[x] < L[x]) ok = 0;
  t[y] = x;
}
int main() {
  scanf("%d", &N);
  if (N == 1) {
    printf("0\n");
    return 0;
  }
  for (int i = 1; i < N; i++)
    scanf("%d %d %d", &edge[i].second.first, &edge[i].second.second,
          &edge[i].first);
  sort(edge + 1, edge + N);
  for (int i = 1; i <= N; i++) {
    scanf("%d", &mx[i]);
    t[i] = i;
    L[i] = 1;
    tot += mx[i];
  }
  ok = 1;
  int ans = edge[1].first;
  for (int i = 1; i < N;) {
    int j = i;
    while (edge[j].first == edge[i].first && j < N)
      unite(edge[j].second.first, edge[j].second.second), j++;
    if (ok) ans = edge[j].first;
    i = j;
  }
  printf("%d\n", ans);
  return 0;
}