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;
struct vvert {
  int u, v, w;
  void read() { scanf("%d %d %d", &u, &v, &w); }
  bool operator<(const vvert &e) const { return w < e.w; }
};
int N, X[3010];
vvert vverts[3010];
int rent[3010], skunk[3010], zeta[3010];
int dara(int x) { return x == rent[x] ? x : rent[x] = dara(rent[x]); }
int main() {
  scanf("%d", &N);
  for (int i = 0; i < N - 1; i++) {
    vverts[i].read();
  }
  sort(vverts, vverts + (N - 1));
  for (int i = 1; i <= N; i++) {
    scanf("%d", &X[i]);
    rent[i] = i;
  }
  copy(X + 1, X + N + 1, skunk + 1);
  fill(zeta + 1, zeta + N + 1, 1);
  int skunkx = accumulate(X + 1, X + N + 1, 0);
  for (int i = 0; i < N - 1; i++) {
    int a = dara(vverts[i].u), b = dara(vverts[i].v);
    if (a != b) {
      rent[a] = b;
      zeta[b] += zeta[a];
      skunk[b] += skunk[a];
      if (zeta[b] > skunkx - skunk[b]) {
        printf("%d\n", vverts[i].w);
        return 0;
      }
    }
  }
  printf("0");
}