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;
template <class T>
void debug(T a, T b) {
  ;
}
template <class T>
void chmin(T& a, const T& b) {
  if (a > b) a = b;
}
template <class T>
void chmax(T& a, const T& b) {
  if (a < b) a = b;
}
namespace std {
template <class S, class T>
ostream& operator<<(ostream& out, const pair<S, T>& a) {
  out << '(' << a.first << ',' << a.second << ')';
  return out;
}
}  // namespace std
int n;
pair<int, pair<int, int> > es[3005];
vector<pair<int, int> > g[3005];
int x[3005];
struct uf {
  static const int MSIZE = 3005;
  int par[MSIZE];
  int size[MSIZE];
  int cap[MSIZE];
  void init() {
    memset(par, -1, sizeof(par));
    for (int i = 0; i < (MSIZE); ++i) {
      size[i] = 1;
      cap[i] = x[i];
    }
  }
  int root(int a) {
    if (par[a] == -1) return a;
    return par[a] = root(par[a]);
  }
  void unite(int a, int b) {
    a = root(a);
    b = root(b);
    if (a == b) return;
    if (size[a] < size[b]) swap(a, b);
    par[b] = a;
    size[a] += size[b];
    cap[a] += cap[b];
  }
  bool same(int a, int b) { return root(a) == root(b); }
};
uf u;
bool check(int w) {
  u.init();
  for (int i = 0; i < (n); ++i)
    if (es[i].first < w) {
      u.unite(es[i].second.first, es[i].second.second);
    }
  vector<pair<int, int> > trees;
  for (int i = 0; i < (n); ++i)
    if (u.root(i) == i) {
      trees.push_back(make_pair(u.size[i], u.cap[i]));
    }
  sort((trees).begin(), (trees).end());
  reverse((trees).begin(), (trees).end());
  ;
  ;
  ;
  ;
  ;
  ;
  if (trees[0].first * 2 > n) {
    int othersum = 0;
    for (int i = (1); i < (trees.size()); ++i) othersum += trees[i].second;
    if (othersum >= trees[0].first) return true;
    return false;
  } else
    return true;
}
int main() {
  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));
    es[i] = make_pair(c, make_pair(a, b));
  }
  for (int i = 0; i < (n); ++i) cin >> x[i];
  int lb = 0, ub = 10005;
  while (ub - lb > 1) {
    int md = (lb + ub) >> 1;
    if (check(md))
      lb = md;
    else
      ub = md;
  }
  cout << lb << endl;
  return 0;
}