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 N = 3e3 + 77, Mod = 1e9 + 7;
int n, p[N], x[N], A, sum;
int find(int v) { return p[v] < 0 ? v : (p[v] = find(p[v])); }
bool merge(int v, int u) {
  v = find(v);
  u = find(u);
  p[v] += p[u];
  x[v] += x[u];
  p[u] = v;
  return sum - x[v] < -p[v];
}
vector<pair<int, pair<int, int> > > E;
int main() {
  fill(p, p + N, -1);
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i < n; i++) {
    int v, u, w;
    cin >> v >> u >> w;
    E.push_back(make_pair(w, make_pair(v, u)));
  }
  for (int i = 1; i <= n; i++) {
    cin >> x[i];
    sum += x[i];
  }
  sort(E.begin(), E.end());
  for (auto s : E) {
    int u = find(s.second.first), v = find(s.second.second), w = s.first;
    A = w;
    if (merge(u, v)) break;
  }
  cout << A;
}