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 = 3001;
int ex[MAXN], ey[MAXN], w[MAXN], f[MAXN], sum, par[MAXN], sz[MAXN], sumF[MAXN];
pair<int, int> all[MAXN];
bool bad;
int find(int x) {
  if (par[x] != x) par[x] = find(par[x]);
  return par[x];
}
void merge(int x, int y) {
  x = find(x);
  y = find(y);
  if (x == y) return;
  sz[x] += sz[y];
  sumF[x] += sumF[y];
  par[y] = x;
  if (sum - sumF[x] < sz[x]) bad = true;
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    cin >> ex[i] >> ey[i] >> w[i];
    ex[i]--;
    ey[i]--;
    all[i] = pair<int, int>(w[i], i);
  }
  for (int i = 0; i < n; i++)
    cin >> f[i], sum += f[i], sumF[i] = f[i], par[i] = i, sz[i] = 1;
  sort(all, all + n - 1);
  int ans = 0, cur = 0;
  while (cur < n - 1) {
    int edgeId = all[cur].second;
    if (!bad) ans = w[edgeId];
    merge(ex[edgeId], ey[edgeId]);
    while (cur < n - 1 && w[edgeId] == all[cur].first)
      merge(ex[all[cur].second], ey[all[cur].second]), cur++;
  }
  if (!bad && n > 1) ans = w[all[n - 2].second];
  cout << ans << endl;
  return 0;
}