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;
int n, par[((int)21 * 1000)], all, a[((int)21 * 1000)], b[((int)21 * 1000)],
    c[((int)21 * 1000)];
vector<pair<int, pair<int, int> > > ed;
int root(int x) {
  if (par[x] == -1) return x;
  return par[x] = root(par[x]);
}
bool check(int x) {
  memset(par, -1, sizeof par);
  for (auto q : ed) {
    if (q.first >= x) break;
    par[root(q.second.first)] = root(q.second.second);
  }
  memset(b, 0, sizeof b);
  memset(c, 0, sizeof c);
  for (int i = 1; i <= n; i++) b[root(i)]++, c[root(i)] += a[i];
  for (int i = 1; i <= n; i++)
    if (par[i] == -1 && all - c[i] < b[i]) return 0;
  return 1;
}
int main() {
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int v, u, w;
    cin >> v >> u >> w;
    ed.push_back({w, {v, u}});
  }
  for (int i = 1; i <= n; i++) cin >> a[i], all += a[i];
  sort(ed.begin(), ed.end());
  int l = 0, r = 20 * 1000;
  while (l < r - 1) {
    int mid = l + r >> 1;
    if (check(mid))
      l = mid;
    else
      r = mid;
  }
  cout << l;
  return 0;
}