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 size[3010], fa[3010], sum[3010], head[10010];
struct node {
  int x, y, next;
} e[3010];
int find(int x) {
  if (x != fa[x]) fa[x] = find(fa[x]);
  return fa[x];
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int x, y, w;
    scanf("%d%d%d", &x, &y, &w);
    e[i].x = x;
    e[i].y = y;
    e[i].next = head[w];
    head[w] = i;
  }
  int s = 0;
  for (int i = 1; i <= n; i++) {
    fa[i] = i;
    size[i] = 1;
    scanf("%d", &sum[i]);
    s += sum[i];
  }
  for (int i = 1; i <= 10000; i++) {
    int p = head[i];
    while (p) {
      int x = find(e[p].x), y = find(e[p].y);
      fa[y] = x;
      size[x] += size[y];
      sum[x] += sum[y];
      if (size[x] + sum[x] > s) {
        printf("%d\n", i);
        return 0;
      }
      p = e[p].next;
    }
  }
  puts("0");
  return 0;
}