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>
inline void umax(T &x, T y) {
  if (y > x) x = y;
}
template <class T>
inline void umin(T &x, T y) {
  if (y < x) x = y;
}
const int N = 2e5 + 5, B = 0x7fffffff;
pair<int, pair<int, int> > edge[N];
int cap[N], siz[N], root[N];
int n;
int pq;
int ans;
int finds(int x) {
  if (root[x] == x) return x;
  return root[x] = finds(root[x]);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++)
    scanf("%d %d %d", &edge[i].second.first, &edge[i].second.second,
          &edge[i].first);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &cap[i]);
    siz[i] = 1;
    root[i] = i;
    pq += cap[i];
  }
  sort(edge + 1, edge + n);
  for (int i = 1; i <= n; i++) {
    int x = finds(edge[i].second.first);
    int y = finds(edge[i].second.second);
    siz[x] += siz[y];
    cap[x] += cap[y];
    root[y] = root[x];
    if (edge[i].first == edge[i + 1].first) continue;
    if (siz[x] > pq - cap[x]) {
      ans = i;
      break;
    }
  }
  printf("%d\n", max(0, edge[ans].first));
  return !!0;
}