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 = 3000 + 5;
struct edge {
  int u, v, a;
  bool operator<(const edge& c) const { return a < c.a; }
} Set[maxn];
int f[maxn], sz[maxn], out[maxn], sum, N;
int F(int o) { return o == f[o] ? o : f[o] = F(f[o]); }
int main() {
  scanf("%d", &N);
  if (N == 1) {
    puts("0");
    return 0;
  }
  for (int i = 1; i < N; ++i) scanf("%d%d%d", &Set[i].u, &Set[i].v, &Set[i].a);
  for (int i = 1; i <= N; ++i) {
    scanf("%d", &sz[i]);
    sum += sz[i];
    out[i] = 1;
    f[i] = i;
  }
  sort(Set + 1, Set + N);
  for (int i = 1; i < N; i++) {
    int u = F(Set[i].u), v = F(Set[i].v);
    if (u != v) {
      f[v] = u;
      sz[u] += sz[v];
      out[u] += out[v];
      if (sz[u] + out[u] > sum) {
        printf("%d\n", Set[i].a);
        return 0;
      }
    }
  }
}