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;
int n, ds[3001], sz[3001], ot[3001], a, b, c, no, ans, sum;
vector<pair<int, pair<int, int> > > v;
int get(int x) {
  if (ds[x] == x)
    return x;
  else
    return ds[x] = get(ds[x]);
}
void merge(int x, int y) {
  sz[x] += sz[y];
  ds[y] = x;
  ot[x] += ot[y];
  if (sz[x] > sum - ot[x]) no = 1;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n - 1; ++i) {
    cin >> a >> b >> c;
    v.push_back(make_pair(c, make_pair(a, b)));
  }
  sort(v.begin(), v.end());
  for (int i = 1; i <= n; ++i) {
    cin >> ot[i];
    sz[i] = 1;
    ds[i] = i;
    sum += ot[i];
  }
  for (int i = 1; i <= n; ++i)
    if (sz[i] > sum - ot[i]) no = 1;
  if (!no && n > 1) ans = v[0].first;
  for (int i = 0; i < v.size();) {
    int j = i;
    while (j < v.size() && v[i].first == v[j].first) {
      merge(get(v[j].second.first), get(v[j].second.second));
      ++j;
    }
    i = j;
    if (!no && i < v.size()) ans = v[i].first;
  }
  cout << ans << '\n';
}