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 = 3e3 + 20;
int x[maxn], from[maxn], to[maxn], w[maxn], n, sx, weight, totalx, tver;
vector<int> adj[maxn];
bool visited[maxn];
void dfs(int v) {
  totalx += x[v];
  tver++;
  visited[v] = 1;
  for (auto e : adj[v]) {
    if (w[e] >= weight) continue;
    int u = from[e] + to[e] - v;
    if (!visited[u]) dfs(u);
  }
}
bool check(int x) {
  weight = x;
  memset(visited, 0, sizeof visited);
  for (int i = 1; i <= n; i++)
    if (!visited[i]) {
      totalx = 0;
      tver = 0;
      dfs(i);
      if (sx - totalx < tver) return 0;
    }
  return 1;
}
int main() {
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    cin >> from[i] >> to[i] >> w[i];
    adj[from[i]].push_back(i);
    adj[to[i]].push_back(i);
  }
  for (int i = 1; i <= n; i++) cin >> x[i], sx += x[i];
  int l = 0, r = 1e4 + 20;
  while (r - l > 1) {
    int m = (r + l) / 2;
    if (check(m))
      l = m;
    else
      r = m;
  }
  cout << l << endl;
}