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;
inline long long read() {
  char ch = getchar();
  long long x = 0, f = 1;
  for (; ch > '9' || ch < '0'; ch = getchar())
    if (ch == '-') f = -1;
  for (; ch >= '0' && ch <= '9'; ch = getchar())
    x = (x << 1) + (x << 3) + ch - '0';
  return x * f;
}
struct bian {
  long long u, v, w;
} e[100000 + 10];
inline bool mmp(bian a, bian b) { return a.w < b.w; }
long long fa[100000 + 10], size[100000 + 10], val[100000 + 10], n, sum;
inline long long find(long long x) {
  return fa[x] == x ? x : fa[x] = find(fa[x]);
}
signed main() {
  n = read();
  for (register long long i = 1; i < n; ++i) {
    e[i].u = read(), e[i].v = read(), e[i].w = read();
  }
  for (register long long i = 1; i <= n; ++i) val[i] = read(), sum += val[i];
  for (register long long i = 1; i <= n; ++i) fa[i] = i, size[i] = 1;
  sort(e + 1, e + n, mmp);
  for (register long long i = 1; i < n; ++i) {
    long long fx = find(e[i].u), fy = find(e[i].v);
    fa[fx] = fy;
    size[fy] += size[fx];
    val[fy] += val[fx];
    if (size[fy] > sum - val[fy]) {
      printf("%lld\n", e[i].w);
      return 0;
    }
  }
  printf("%lld\n", e[n - 1].w);
  return 0;
}