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 a[3000];
int par[3000];
int cnt[3000];
void init(int n) {
  int i;
  for (i = 0; i < n; i++) {
    par[i] = i;
    cnt[i] = 0;
  }
}
int find(int x) {
  if (par[x] == x) return x;
  return par[x] = find(par[x]);
}
void unite(int x, int y) {
  x = find(x);
  y = find(y);
  if (x == y) return;
  par[x] = y;
}
int main() {
  int n, l, r, m, i;
  vector<pair<int, pair<int, int> > > v;
  scanf("%d", &n);
  for (i = 0; i < n - 1; i++) {
    int x, y, z;
    scanf("%d %d %d", &x, &y, &z);
    v.push_back(make_pair(z, make_pair(x - 1, y - 1)));
  }
  sort(v.begin(), v.end());
  for (i = 0; i < n; i++) scanf("%d", &a[i]);
  if (n == 1) {
    puts("0");
    return 0;
  }
  l = 0, r = n - 1, m = (l + r) / 2;
  while (r - l > 1) {
    int c = 0, sum = 0;
    init(n);
    for (i = 0; i < m; i++) unite(v[i].second.first, v[i].second.second);
    for (i = 0; i < n; i++) cnt[find(i)]++;
    for (i = 0; i < n; i++) {
      if (cnt[i] > cnt[c]) c = i;
    }
    for (i = 0; i < n; i++) {
      if (find(i) != c) sum += a[i];
    }
    if (sum >= cnt[c]) {
      l = m;
      m = (l + r) / 2;
    } else {
      r = m;
      m = (l + r) / 2;
    }
  }
  printf("%d\n", v[l].first);
  return 0;
}