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 = 3005;
int n, val[MAXN], tot = 0;
int par[MAXN], sz[MAXN];
pair<int, pair<int, int> > edge[MAXN];
int find(int x) { return x == par[x] ? x : par[x] = find(par[x]); }
void join(int x, int y) { par[find(x)] = find(y); }
int main() {
  ios::sync_with_stdio(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--;
  }
  for (int i = 0; i < n; i++) {
    cin >> val[i];
    tot += val[i];
    par[i] = i;
    sz[i] = 1;
  }
  sort(edge, edge + n - 1);
  for (int i = 0; i < n - 1; i++) {
    int a = edge[i].second.first, b = edge[i].second.second;
    if (find(a) == find(b)) continue;
    sz[find(b)] += sz[find(a)];
    val[find(b)] += val[find(a)];
    join(a, b);
    if (sz[find(a)] + val[find(a)] > tot) {
      cout << edge[i].first << '\n';
      return 0;
    }
  }
  cout << "0\n";
}