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 = 100100;
int n, X[maxn];
struct st {
  int a, b, c;
} e[maxn];
bool cmp(st a, st b) { return a.c < b.c; }
int size[maxn], M[maxn], sum, fa[maxn];
int find(int x) { return fa[x] == x ? x : fa[x] = find(fa[x]); }
bool check(int limit) {
  for (int i = 1; i <= n; i++) fa[i] = i, size[i] = 1, M[i] = X[i];
  for (int i = 1; i < n && e[i].c < limit; i++) {
    int f1 = find(e[i].a), f2 = find(e[i].b);
    if (f1 == f2) continue;
    fa[f1] = f2;
    M[f2] += M[f1];
    size[f2] += size[f1];
  }
  for (int i = 1; i <= n; i++) {
    int f = find(i);
    if (size[f] > sum - M[f]) return 0;
  }
  return 1;
}
int main() {
  scanf("%d", &n);
  int mx = 0;
  for (int i = 1; i < n; i++)
    scanf("%d%d%d", &e[i].a, &e[i].b, &e[i].c), mx = max(mx, e[i].c);
  for (int i = 1; i <= n; i++) scanf("%d", &X[i]), sum += X[i];
  int l = 0, r = mx, ans = 0;
  sort(e + 1, e + n, cmp);
  while (l <= r) {
    int mid = (l + r) / 2;
    if (check(mid))
      l = mid + 1, ans = mid;
    else
      r = mid - 1;
  }
  printf("%d", ans);
  return 0;
}