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;
vector<pair<int, int> > G[3000];
int x[3000];
int n;
bool mark[3000];
pair<int, int> operator+(pair<int, int> a, pair<int, int> b) {
  return make_pair(a.first + b.first, a.second + b.second);
}
pair<int, int> dfs(int v, int d) {
  if (mark[v]) return make_pair(0, 0);
  pair<int, int> as(1, x[v]);
  mark[v] = true;
  for (int i = 0; i < G[v].size(); i++)
    if (G[v][i].second < d) as = as + dfs(G[v][i].first, d);
  return as;
}
bool check(int l) {
  vector<pair<int, int> > vv;
  memset(mark, 0, sizeof mark);
  for (int i = 0; i < n; i++) {
    pair<int, int> p = dfs(i, l);
    if (p != make_pair(0, 0)) vv.push_back(p);
  }
  int sm = 0;
  for (int i = 0; i < vv.size(); i++) sm += vv[i].second;
  for (int i = 0; i < vv.size(); i++) {
    if (sm - vv[i].second < vv[i].first) {
      return false;
    }
  }
  return true;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int a, b, c;
    cin >> a >> b >> c;
    a--;
    b--;
    G[a].push_back(make_pair(b, c));
    G[b].push_back(make_pair(a, c));
  }
  for (int i = 0; i < n; i++) cin >> x[i];
  int l = 0, r = 100000;
  while (l + 1 < r) {
    int m = (l + r) / 2;
    if (check(m))
      l = m;
    else
      r = m;
  }
  cout << l << endl;
}