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;
constexpr int N = 3000 + 10;
constexpr int LG = 20;
constexpr int MOD = 1e9 + 7;
constexpr int MOD2 = 1e9 + 9;
int n, par[N], u, v, w, sz[N], sum[N], ans, sumall;
vector<pair<int, pair<int, int>>> e;
int get_par(int v) {
  if (par[v] == v) return v;
  return par[v] = get_par(par[v]);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i < n; i++) {
    cin >> u >> v >> w;
    e.push_back({w, {u, v}});
  }
  for (int i = 1; i <= n; i++)
    cin >> sum[i], par[i] = i, sz[i] = 1, sumall += sum[i];
  sort(e.begin(), e.end());
  for (int i = 0; i < (int)e.size(); i++) {
    u = e[i].second.first;
    v = e[i].second.second;
    par[u] = get_par(u);
    par[v] = get_par(v);
    sum[par[u]] += sum[par[v]];
    sz[par[u]] += sz[par[v]];
    par[par[v]] = par[u];
    ans = e[i].first;
    if (sz[par[u]] >= (n + 1) / 2)
      if (sz[par[u]] > sumall - sum[par[u]]) break;
  }
  cout << ans << '\n';
  return 0;
}