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;
int N;
pair<int, pair<int, int> > E[3000];
int ans, sum;
bool ok = true;
struct UnionFind {
  int par[3000], ra[3000], sz[3000];
  void init() {
    for (int(i) = 0; (i) < (3000); (i)++) par[i] = i, ra[i] = 1;
  }
  int find(int x) { return par[x] == x ? x : par[x] = find(par[x]); }
  void unite(int a, int b) {
    if ((a = find(a)) != (b = find(b))) {
      if (ra[a] < ra[b]) swap(a, b);
      par[b] = a, ra[a] += ra[b];
      sz[a] += sz[b];
      if (sum - sz[a] < ra[a]) ok = false;
    }
  }
} uf;
int main() {
  scanf("%d", &N);
  for (int(i) = 0; (i) < (N - 1); (i)++)
    scanf("%d%d%d", &E[i].second.first, &E[i].second.second, &E[i].first),
        E[i].second.first--, E[i].second.second--;
  sort(E, E + N - 1);
  uf.init();
  sum = 0;
  for (int(i) = 0; (i) < (N); (i)++) scanf("%d", &uf.sz[i]), sum += uf.sz[i];
  for (int(i) = 0; (i) < (N); (i)++) ok &= (sum - uf.sz[i] >= 1);
  if (ok) ans = E[0].first;
  if (N == 1) {
    puts("0");
    return 0;
  }
  int ptr = 0;
  while (ptr < N - 1) {
    int tt = E[ptr].first;
    while (ptr < N - 1 && tt == E[ptr].first) {
      uf.unite(E[ptr].second.first, E[ptr].second.second);
      ptr++;
    }
    if (ok && ptr <= N - 1) ans = E[min(ptr, N - 2)].first;
  }
  printf("%d\n", ans);
}