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 int M = 1e4 + 10, N = 3100;
vector<pair<pair<int, int>, int> > ed;
int n, par[N], sz[N], sig[N], x[N];
int root(int v) {
  if (par[v] == -1) return v;
  return par[v] = root(par[v]);
}
void merge(int u, int v) {
  u = root(u), v = root(v);
  if (u == v) return;
  par[u] = v;
}
bool check(int val) {
  memset(par, -1, sizeof par);
  memset(sz, 0, sizeof sz);
  memset(sig, 0, sizeof sig);
  for (int i = 0; i < n - 1; i++) {
    int v = ed[i].first.first, u = ed[i].first.second, w = ed[i].second;
    if (w < val) merge(u, v);
  }
  for (int i = 1; i <= n; i++) {
    sz[root(i)]++;
    sig[root(i)] += x[i];
  }
  int SIG = 0;
  for (int i = 1; i <= n; i++) {
    if (par[i] == -1) SIG += sig[i];
  }
  for (int i = 1; i <= n; i++) {
    if (sz[root(i)] > SIG - sig[root(i)]) return false;
  }
  return true;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int a, b, c;
    cin >> a >> b >> c;
    ed.push_back({{a, b}, c});
  }
  for (int i = 1; i <= n; i++) cin >> x[i];
  int lo = 0, hi = M;
  while (hi - lo > 1) {
    int mid = (lo + hi) / 2;
    if (check(mid))
      lo = mid;
    else
      hi = mid;
  }
  cout << lo << '\n';
}