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;
template <class T>
T abs(T x) {
  return x > 0 ? x : -x;
}
int n;
int m;
int p[10000];
int x[10000];
int y[10000];
int w[10000];
pair<int, pair<int, int> > v[10000];
int gp(int x) {
  if (p[x] == x) return x;
  return p[x] = gp(p[x]);
}
int can(int h) {
  for (int i = 0; i < n; i++) {
    p[i] = i;
    x[i] = 1;
    y[i] = w[i];
  }
  for (int i = 0; i < h; i++) {
    int a = gp(v[i].second.first);
    int b = gp(v[i].second.second);
    if (a != b) {
      p[a] = b;
      x[b] += x[a];
      y[b] += y[a];
    }
  }
  int sx = 0, sy = 0;
  vector<pair<int, int> > u;
  for (int i = 0; i < n; i++)
    if (p[i] == i) {
      u.push_back(make_pair(x[i], y[i]));
      sx += x[i];
      sy += y[i];
    }
  for (int i = 0; i < ((int)(u).size()); i++)
    if (u[i].first > sy - u[i].second) return 0;
  return 1;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i + 1 < n; i++) {
    int a, b, c;
    scanf("%d%d%d", &a, &b, &c);
    a--;
    b--;
    v[i] = make_pair(c, make_pair(a, b));
  }
  for (int i = 0; i < n; i++) scanf("%d", &w[i]);
  sort(v, v + n - 1);
  int l = 0, r = n;
  while (r - l > 1) {
    int s = (l + r) / 2;
    if (can(s))
      l = s;
    else
      r = s;
  }
  printf("%d\n", v[l].first);
  return 0;
}