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 double eps(1e-8);
const double pi(3.14159265358979);
const int N = 3030;
struct edge {
  int u, v, l;
} e[N] = {};
bool edge_cmp(const edge &e1, const edge &e2) { return e1.l < e2.l; }
int n, root[N] = {}, s1[N] = {}, s2[N] = {}, sum1 = 0, sum2 = 0;
int get_root(int r) {
  if (r != root[r]) root[r] = get_root(root[r]);
  return root[r];
}
void print_root() {
  for (int i = 1; i <= n; ++i) get_root(i);
  for (int i = 1; i <= n; ++i) cout << root[i] << ' ';
  cout << endl;
  for (int i = 1; i <= n; ++i) cout << s1[i] << ' ';
  cout << endl;
  for (int i = 1; i <= n; ++i) cout << s2[i] << ' ';
  cout << endl;
}
void init() {
  scanf("%d", &n);
  for (int i = 1; i < n; ++i) scanf("%d%d%d", &e[i].u, &e[i].v, &e[i].l);
  sort(e + 1, e + n, edge_cmp);
  for (int i = 1; i <= n; ++i) root[i] = i, s1[i] = 1, scanf("%d", &s2[i]);
  sum1 = n;
  sum2 = accumulate(s2 + 1, s2 + n + 1, 0);
}
void work() {
  int i = 0;
  for (i = 1; i < n; ++i) {
    int u = get_root(e[i].u), v = get_root(e[i].v);
    root[v] = u;
    s1[u] += s1[v];
    s2[u] += s2[v];
    if (s1[u] > sum2 - s2[u]) break;
  }
  cout << e[i].l << endl;
}
int main() {
  init();
  work();
  return 0;
}