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;
struct st {
  int sz, cp, pr;
} data[1 << 15];
int n;
int a, b, c, sm, ans;
vector<pair<int, pair<int, int>>> E(1);
int P(int x) {
  if (data[x].pr == x) return x;
  return data[x].pr = P(data[x].pr);
}
int main() {
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) {
    scanf("%d%d%d", &a, &b, &c);
    E.push_back({c, {a, b}});
  }
  sort(E.begin(), E.end());
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a);
    sm += a;
    data[i].pr = i;
    data[i].cp = a;
    data[i].sz = 1;
  }
  for (int i = 1; i <= n - 1; i++) {
    a = P(E[i].second.first);
    b = P(E[i].second.second);
    data[b].pr = a;
    data[a].sz += data[b].sz;
    data[a].cp += data[b].cp;
    ans = i;
    if (data[a].sz > sm - data[a].cp) break;
  }
  printf("%d", E[ans].first);
}