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 MAXN = 3005;
int n, x[MAXN];
vector<pair<int, int> > adj[MAXN];
int t, sze, num, sum;
bool vis[MAXN];
void dfs(int cur) {
  if (vis[cur]) return;
  vis[cur] = 1;
  sze++;
  num += x[cur];
  for (int i = 0; i < adj[cur].size(); i++)
    if (adj[cur][i].second < t) dfs(adj[cur][i].first);
}
bool check(int mid) {
  t = mid;
  memset(vis, 0, sizeof(vis));
  for (int i = 0; i < n; i++)
    if (!vis[i]) {
      sze = num = 0;
      dfs(i);
      if (sze > sum - num) return 0;
    }
  return 1;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  if (fopen("input.txt", "r")) freopen("input.txt", "r", stdin);
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int a, b, c;
    cin >> a >> b >> c;
    a--, b--;
    adj[a].push_back(pair<int, int>(b, c));
    adj[b].push_back(pair<int, int>(a, c));
  }
  for (int i = 0; i < n; i++) {
    cin >> x[i];
    sum += x[i];
  }
  int lo = 0, hi = 10000;
  while (lo < hi) {
    int mid = (lo + hi + 1) / 2;
    if (check(mid))
      lo = mid;
    else
      hi = mid - 1;
  }
  cout << lo << '\n';
  return 0;
}