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 <typename T, typename U>
ostream& operator<<(ostream& os, const pair<T, U>& p) {
  os << "(" << p.first << "," << p.second << ")";
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  os << "[";
  for (__typeof((v).begin()) i = (v).begin(); i != (v).end(); ++i) {
    if (i != v.begin()) os << ", ";
    os << *i;
  }
  os << "]";
  return os;
}
vector<int> dad;
vector<int> X;
int find_set(int x) { return dad[x] < 0 ? x : dad[x] = find_set(dad[x]); }
void union_sets(int x, int y) {
  x = find_set(x);
  y = find_set(y);
  if (x == y) return;
  if (dad[x] > dad[y]) swap(x, y);
  dad[x] += dad[y];
  X[x] += X[y];
  dad[y] = x;
}
int main() {
  int n;
  scanf("%d", &n);
  vector<pair<int, pair<int, int> > > E(n - 1);
  for (int i = 0; i < (int)(n - 1); ++i) {
    scanf("%d%d%d", &E[i].second.first, &E[i].second.second, &E[i].first);
    --E[i].second.first;
    --E[i].second.second;
  }
  X.resize(n);
  for (int i = 0; i < (int)(n); ++i) scanf("%d", &X[i]);
  dad.resize(n, -1);
  int C = accumulate(X.begin(), X.end(), 0);
  sort(E.begin(), E.end());
  int res = 0;
  for (__typeof((E).begin()) e = (E).begin(); e != (E).end(); ++e) {
    union_sets(e->second.first, e->second.second);
    int i = find_set(e->second.first);
    if (-dad[i] > C - X[i]) {
      res = e->first;
      break;
    }
  }
  printf("%d\n", res);
}