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 inf = 1000000000;
int sz[50005], f[50005], b[50005];
int sum;
pair<int, pair<int, int> > a[5005];
bool isOk;
int n;
int find_set(int x) {
  if (f[x] == x) return x;
  f[x] = find_set(f[x]);
  return f[x];
}
void unite(int x, int y) {
  x = find_set(x);
  y = find_set(y);
  if (x == y) return;
  f[x] = y;
  b[y] += b[x];
  sz[y] += sz[x];
  if (sz[y] > sum - b[y]) isOk = false;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n - 1; ++i) {
    scanf("%d%d%d", &a[i].second.first, &a[i].second.second, &a[i].first);
    a[i].second.first--;
    a[i].second.second--;
  }
  sort(a, a + n - 1);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &b[i]);
    sum += b[i];
    sz[i] = 1;
  }
  for (int i = 0; i < n; ++i) f[i] = i;
  isOk = true;
  int ans = a[0].first;
  for (int i = 0; i < n; ++i) {
    if (sz[i] > sum - b[i]) isOk = false;
  }
  for (int i = 0; i < n - 1;) {
    int j = i;
    while (j < n - 1 && a[i].first == a[j].first) {
      unite(a[j].second.first, a[j].second.second);
      j++;
    }
    if (!isOk) break;
    if (a[j].first != 0)
      ans = a[j].first;
    else
      ans = a[j - 1].first;
    i = j;
  }
  printf("%d\n", ans);
  return 0;
}