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 MAX_N = 3e3 + 9;
int n, ans;
pair<int, pair<int, int> > edge[MAX_N];
int par[MAX_N];
int len[MAX_N];
int val[MAX_N];
int sum;
int root(int v) { return par[v] == v ? v : par[v] = root(par[v]); }
int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    cin >> edge[i].second.first >> edge[i].second.second >> edge[i].first;
    edge[i].second.first--, edge[i].second.second--;
  }
  sort(edge, edge + n - 1);
  for (int i = 0; i < n; i++) {
    par[i] = i;
    len[i] = 1;
    cin >> val[i];
    sum += val[i];
  }
  ans = edge[0].first;
  for (int i = 0; i < n; i++) {
    int v = edge[i].second.first;
    int u = edge[i].second.second;
    val[root(u)] += val[root(v)];
    len[root(u)] += len[root(v)];
    par[root(v)] = root(u);
    if (sum - val[root(v)] < len[root(v)]) break;
    ans = edge[i + 1].first;
  }
  cout << ans << "\n";
  return 0;
}