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 pa[5009], siz[5009], occ[5009], tots = 0;
bool yay;
vector<pair<int, pair<int, int> > > edges;
int finder(int node) {
  if (pa[node] != node) {
    return pa[node] = finder(pa[node]);
  }
  return node;
}
void uniter(int x, int y) {
  x = finder(x);
  y = finder(y);
  if (x == y) {
    return;
  }
  pa[y] = x;
  siz[x] += siz[y];
  occ[x] += occ[y];
  if (tots - occ[x] < siz[x]) {
    yay = false;
  }
  return;
}
int main() {
  int i, j, t1, t2, t3, t4, n, ans = 0;
  scanf("%d", &n);
  for (i = 0; i < n - 1; i++) {
    scanf("%d %d %d", &t1, &t2, &t3);
    edges.push_back(make_pair(t3, make_pair(t1, t2)));
  }
  sort(edges.begin(), edges.end());
  for (i = 1; i <= n; i++) {
    pa[i] = i;
    siz[i] = 1;
    scanf("%d", &occ[i]);
    tots += occ[i];
  }
  if (n == 1) {
    printf("0\n");
    return 0;
  }
  ans = edges[0].first;
  yay = true;
  for (i = 0; i < n - 1;) {
    j = i;
    while (edges[i].first == edges[j].first && j < n - 1) {
      uniter(edges[j].second.first, edges[j].second.second);
      j++;
    }
    if (yay && j < n - 1) {
      ans = edges[j].first;
    }
    i = j;
  }
  printf("%d\n", ans);
  return 0;
}