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 MAXN = 1e5 + 10;
inline int read() {
  char c = getchar();
  int x = 0, f = 1;
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
  return x * f;
}
int N, fa[MAXN], x[MAXN], siz[MAXN], sum;
struct Edge {
  int u, v, w;
  bool operator<(const Edge &rhs) const { return w < rhs.w; }
} E[MAXN];
int find(int x) { return fa[x] == x ? x : fa[x] = find(fa[x]); }
int main() {
  N = read();
  for (int i = 1; i <= N; i++) fa[i] = i, siz[i] = 1;
  for (int i = 1; i <= N - 1; i++)
    E[i].u = read(), E[i].v = read(), E[i].w = read();
  sort(E + 1, E + N);
  for (int i = 1; i <= N; i++) x[i] = read(), sum += x[i];
  for (int i = 1; i <= N - 1; i++) {
    int fx = find(E[i].u), fy = find(E[i].v);
    if (fx == fy) continue;
    fa[fy] = fx;
    siz[fx] += siz[fy];
    x[fx] += x[fy];
    if (siz[fx] > sum - x[fx]) {
      printf("%d\n", E[i].w);
      return 0;
    }
  }
  printf("%d\n", E[N - 1].w);
  return 0;
}