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 = 3005;
int n, ans, can, all;
int f[maxn], siz[maxn], sum[maxn];
pair<int, pair<int, int> > E[maxn];
int getf(int x) { return f[x] ? f[x] = getf(f[x]) : x; }
void merge(int x, int y) {
  x = getf(x), y = getf(y);
  f[y] = x;
  siz[x] += siz[y];
  sum[x] += sum[y];
  if (siz[x] > all - sum[x]) can = 0;
}
void init() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++)
    scanf("%d%d%d", &E[i].second.first, &E[i].second.second, &E[i].first);
  sort(E + 1, E + n);
  for (int i = 1; i <= n; i++) siz[i] = 1, scanf("%d", sum + i), all += sum[i];
}
int work() {
  if (n == 1) return 0;
  can = 1;
  for (int i = 1; i <= n; i++)
    if (siz[i] > all - sum[i]) can = 0;
  if (can) ans = E[1].first;
  for (int i = 1, j; i < n && can; i = j) {
    for (j = i; j < n && E[j].first == E[i].first; j++)
      merge(E[j].second.first, E[j].second.second);
    if (can && j < n) ans = E[j].first;
  }
  return ans;
}
int main() {
  init();
  cout << work() << endl;
  return 0;
}