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 int N = 3010;
int i, j, n, fa[N], sum[N], num[N], tot;
struct E {
  int u, v, w;
  bool operator<(const E& a) const { return w < a.w; }
} e[N];
int get(int x) {
  if (fa[x] == x) return x;
  return fa[x] = get(fa[x]);
}
int main() {
  scanf("%d", &n);
  for (i = 1; i < n; i++) scanf("%d %d %d", &e[i].u, &e[i].v, &e[i].w);
  for (i = 1; i <= n; i++) scanf("%d", sum + i);
  for (i = 1; i <= n; i++) fa[i] = i, num[i] = 1, tot += sum[i];
  sort(e + 1, e + n);
  for (i = 1; i < n; i++) {
    static int x, y;
    for (j = 1; j <= n; j++)
      if (fa[j] == j && num[j] > n / 2) break;
    if (num[j] > tot - sum[j]) break;
    x = get(e[i].u), y = get(e[i].v);
    if (x != y) {
      num[y] += num[x];
      sum[y] += sum[x];
      fa[x] = y;
    }
  }
  cout << e[i - 1].w << endl;
}