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 INF = 1e9 + 10;
const int MN = 3e3 + 10;
int es[MN], ee[MN], el[MN], f[MN], sum, par[MN], sz[MN], F[MN];
pair<int, int> borobaba[MN];
bool fl;
int n;
int root(int x) {
  if (par[x] == x) return x;
  return par[x] = root(par[x]);
}
void Union(int x, int y) {
  x = root(x), y = root(y);
  if (x == y) return;
  if (sz[x] < sz[y]) swap(x, y);
  sz[x] += sz[y];
  F[x] += F[y];
  par[y] = x;
  if (sum - F[x] < sz[x]) fl = true;
}
int main() {
  cin >> n;
  for (int i = 0; i < n - 1; ++i) {
    cin >> es[i] >> ee[i] >> el[i];
    --es[i];
    --ee[i];
    borobaba[i] = pair<int, int>(el[i], i);
  }
  for (int i = 0; i < n; ++i)
    cin >> f[i], sum += f[i], F[i] = f[i], par[i] = i, sz[i] = 1;
  sort(borobaba, borobaba + n - 1);
  int ans = 0, cur = 0;
  while (cur < n - 1) {
    int Id = borobaba[cur].second;
    if (!fl) ans = el[Id];
    Union(es[Id], ee[Id]);
    while (cur < n - 1 && el[Id] == borobaba[cur].first)
      Union(es[borobaba[cur].second], ee[borobaba[cur].second]), cur++;
  }
  if (!fl && n > 1) ans = el[borobaba[n - 2].second];
  cout << ans << endl;
  return 0;
}