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() {
  long long nm = 0;
  bool fh = true;
  char cw = getchar();
  for (; !isdigit(cw); cw = getchar()) fh ^= (cw == '-');
  for (; isdigit(cw); cw = getchar()) nm = nm * 10 + (cw - '0');
  return fh ? nm : -nm;
}
int n, hv[200200], all, F[200200], sz[200200];
struct Edge {
  int x, y, z;
  inline void gtin() { x = read(), y = read(), z = read(); }
  inline bool operator<(const Edge b) const { return z < b.z; }
} e[200200];
inline int fd(int x) { return (F[x] == x) ? x : (F[x] = fd(F[x])); }
int main() {
  n = read();
  for (int i = 1; i < n; i++) e[i].gtin();
  for (int i = 1; i <= n; i++)
    hv[i] = read(), all += hv[i], F[i] = i, sz[i] = 1;
  sort(e + 1, e + n);
  for (int i = 1, x, y, fx, fy; i < n; i++) {
    x = e[i].x, y = e[i].y, fx = fd(x), fy = fd(y);
    if (fx ^ fy) {
      hv[fx] += hv[fy], sz[fx] += sz[fy], F[fy] = fx;
      if (all - hv[fx] < sz[fx]) {
        printf("%d\n", e[i].z);
        return 0;
      }
    }
  }
  printf("%d\n", e[n - 1].z);
  return 0;
}