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 N = 3000 + 100;
int aa[N], bb[N], cc[N];
int n;
int fa[N], sum[N], son[N], lim[N];
int Find(int u) {
  if (u == fa[u]) return u;
  return fa[u] = Find(fa[u]);
}
int check(int mid) {
  for (int i = 1; i <= n; i++) fa[i] = i, sum[i] = lim[i], son[i] = 1;
  for (int i = 1; i < n; i++) {
    if (cc[i] >= mid) continue;
    int u = Find(aa[i]), v = Find(bb[i]);
    if (u == v) continue;
    if (son[u] > son[v]) swap(u, v);
    fa[u] = v;
    sum[v] += sum[u];
    son[v] += son[u];
  }
  int all = 0;
  for (int i = 1; i <= n; i++) {
    int u = Find(i);
    if (u != i) continue;
    all += sum[i];
  }
  for (int i = 1; i <= n; i++) {
    if (fa[i] != i) continue;
    if (all - sum[i] < son[i]) return 0;
  }
  return 1;
}
int sta[N], top;
int main() {
  while (scanf("%d", &n) != EOF) {
    top = 0;
    for (int i = 1; i < n; i++) {
      scanf("%d%d%d", &aa[i], &bb[i], &cc[i]);
      sta[top++] = cc[i];
    }
    for (int i = 1; i <= n; i++) scanf("%d", &lim[i]);
    sort(sta, sta + top);
    top = unique(sta, sta + top) - sta;
    int l = 1, r = top - 1;
    int ans = sta[0];
    while (l <= r) {
      int mid = (l + r) / 2;
      if (check(sta[mid])) {
        ans = sta[mid];
        l = mid + 1;
      } else
        r = mid - 1;
    }
    printf("%d\n", ans);
  }
  return 0;
}