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;
long w[200000], sz[200000], ttl[200000], ans, n;
vector<pair<long, pair<long, long> > > v;
long s, er, a, b;
long c, cur;
long gett(long x) {
  while (x != w[x]) x = w[x];
  return x;
}
void merg(long a, long b) {
  a = gett(a), b = gett(b);
  w[b] = a;
  sz[a] += sz[b];
  ttl[a] += ttl[b];
  if (sz[a] > s - ttl[a]) er = 1;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int i = 1; i < n; 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 >> ttl[i];
    sz[i] = 1;
    s += ttl[i];
    w[i] = i;
  }
  if (n == 1) {
    cout << 0 << endl;
    return 0;
  }
  ans = v[0].first;
  for (int i = 0; i < v.size();) {
    cur = i;
    while (cur < v.size() && v[cur].first == v[i].first) {
      merg(v[cur].second.first, v[cur].second.second);
      ++cur;
    }
    i = cur;
    if (i < v.size() && er == 0) ans = v[i].first;
  }
  cout << ans << endl;
  cin.get();
  cin.get();
  return 0;
}