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 long double pi = 3.1415926535897932384626433832795l;
template <typename T>
inline auto sqr(T x) -> decltype(x * x) {
  return x * x;
}
template <typename T>
inline T abs(T x) {
  return x > T() ? x : -x;
}
template <typename T1, typename T2>
inline bool umx(T1& a, T2 b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <typename T1, typename T2>
inline bool umn(T1& a, T2 b) {
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}
const int N = 3000;
struct Input {
  int n;
  int a[N], b[N], c[N];
  int x[N];
  void init(const Input& input) { *this = input; }
};
struct Data : Input {
  bool read() {
    if (!(cin >> n)) {
      return 0;
    }
    for (int i = 0; i < int(n - 1); ++i) {
      scanf("%d%d%d", &a[i], &b[i], &c[i]);
      --a[i];
      --b[i];
    }
    for (int i = 0; i < int(n); ++i) {
      scanf("%d", &x[i]);
    }
    return 1;
  }
  int ans;
  void write() { cout << ans << endl; }
  virtual void solve() {}
  virtual void clear() { *this = Data(); }
};
struct Solution : Data {
  int p[N], s[N], t[N];
  int P(int x) {
    if (p[x] == x) {
      return x;
    }
    return p[x] = P(p[x]);
  }
  void uni(int x, int y) {
    x = P(x);
    y = P(y);
    if (s[x] > s[y]) {
      swap(x, y);
    }
    p[x] = y;
    s[y] += s[x];
    t[y] += t[x];
  }
  void solve() {
    int m = accumulate(x, x + n, 0);
    for (int i = 0; i < int(n); ++i) {
      p[i] = i;
      s[i] = 1;
      t[i] = x[i];
    }
    vector<pair<int, pair<int, int> > > v;
    for (int i = 0; i < int(n - 1); ++i) {
      v.push_back(make_pair(c[i], make_pair(a[i], b[i])));
    }
    sort((v).begin(), (v).end());
    for (int i = 0; i < int(n - 1); ++i) {
      uni(v[i].second.first, v[i].second.second);
      int k = P(v[i].second.first);
      if (s[k] > m - t[k]) {
        ans = v[i].first;
        return;
      }
    }
  }
  void clear() { *this = Solution(); }
};
Solution sol;
int main() {
  cout.setf(ios::showpoint | ios::fixed);
  cout.precision(20);
  sol.read();
  sol.solve();
  sol.write();
  return 0;
}