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 = 1e5 + 100;
int gi() {
  char c;
  int w;
  bool q = true;
  while (((c = getchar()) < '0' || '9' < c) && c != '-')
    ;
  if (c == '-')
    q = false, w = 0;
  else
    w = c - '0';
  while ('0' <= (c = getchar()) && c <= '9') w = (w << 1) + (w << 3) + c - '0';
  return q ? w : -w;
}
struct E {
  int u, v, w;
  inline void in() {
    u = gi();
    v = gi();
    w = gi();
  }
  bool operator<(const E &b) const { return w < b.w; }
} e[N];
int fa[N];
inline int find(int x) { return fa[x] == x ? x : fa[x] = find(fa[x]); }
int siz[N], in[N];
int main() {
  int n = gi(), i, a, b, all = 0;
  for (i = 1; i < n; i++) e[i].in();
  for (i = 1; i <= n; i++) siz[i] = 1, all += in[fa[i] = i] = gi();
  sort(e + 1, e + n);
  for (i = 1; i < n; i++) {
    a = find(e[i].u), b = find(e[i].v);
    siz[a] += siz[b], in[a] += in[b], fa[b] = a;
    if (all - in[a] < siz[a]) break;
  }
  printf("%d\n", e[i].w);
}