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;
inline void read(int &q) {
  int x = 0, f = 1;
  char ch;
  for (ch = getchar(); (ch < '0' || ch > '9') && ch != '-'; ch = getchar())
    ;
  if (ch == '-') {
    f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 3) + (x << 1) + ch - '0';
    ch = getchar();
  }
  q = x * f;
}
struct Edge {
  int x, y, w;
} edge[100001];
int n, sum, tot, l, r, mid, flag, Ans;
int father[100001], ans[100001], tmp[100001], size[100001], a[100001],
    x[100001];
int find(int x) { return x == father[x] ? x : father[x] = find(father[x]); }
int main() {
  cin >> n;
  for (int i = 1; i < n; i++) {
    read(edge[i].x);
    read(edge[i].y);
    read(edge[i].w);
    tmp[i] = edge[i].w;
  }
  for (int i = 1; i <= n; i++) {
    read(x[i]);
    tot += x[i];
  }
  sort(tmp + 1, tmp + n);
  sum = 1;
  ans[1] = tmp[1];
  for (int i = 1; i < n; i++)
    if (tmp[i] != tmp[i - 1]) {
      sum++;
      ans[sum] = tmp[i];
    }
  Ans = l = 1;
  r = sum;
  while (l <= r) {
    flag = 1;
    mid = (l + r) >> 1;
    memset(a, 0, sizeof(a));
    memset(size, 0, sizeof(size));
    for (int i = 1; i <= n; i++) father[i] = i;
    for (int i = 1; i < n; i++)
      if (edge[i].w < ans[mid] && find(edge[i].x) != find(edge[i].y))
        father[find(edge[i].x)] = find(edge[i].y);
    for (int i = 1; i <= n; i++) {
      size[find(i)]++;
      a[find(i)] += x[i];
    }
    for (int i = 1; i <= n; i++)
      if (find(i) == i && tot - a[i] < size[i]) {
        flag = 0;
        break;
      }
    if (flag) {
      Ans = max(Ans, mid);
      l = mid + 1;
    } else
      r = mid - 1;
  }
  cout << ans[Ans];
  return 0;
}