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 long long MX = 3009;
long long x[MX], par[MX], num[MX], ans = 0;
pair<long long, pair<long long, long long> > edg[MX];
void MS(long long v) {
  par[v] = -1;
  num[v] = x[v];
}
long long FS(long long v) {
  if (par[v] < 0) return v;
  return par[v] = FS(par[v]);
}
void US(long long a, long long b) {
  a = FS(a), b = FS(b);
  if (a == b) return;
  if (par[a] > par[b]) swap(a, b);
  par[a] += par[b];
  par[b] = a;
  num[a] += num[b];
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long n;
  cin >> n;
  for (long long i = 0; i < n - 1; i++) {
    long long u, v, w;
    cin >> u >> v >> w;
    edg[i] = {w, {u, v}};
  }
  for (long long i = 1; i <= n; i++) cin >> x[i];
  for (long long i = 1; i <= n; i++) MS(i);
  sort(edg, edg + n - 1);
  for (long long i = 0; i < n - 1; i++) {
    long long u = edg[i].second.first, v = edg[i].second.second,
              w = edg[i].first;
    long long maxx = 0, sum = 0, res = 1;
    for (long long j = 1; j <= n; j++) {
      if (par[j] < 0) {
        maxx = max(maxx, -par[j]);
        sum += num[j];
      }
    }
    for (long long j = 1; j <= n; j++) {
      if (par[j] < 0) {
        if (sum - num[j] < -par[j]) res = 0;
      }
    }
    if (res) ans = w;
    US(u, v);
  }
  cout << ans;
  return 0;
}