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;
struct UF {
  int par[3005], rank[3005];
  int cost[3005];
  void init(int n) {
    for (int i = 0; i < n; i++) {
      par[i] = i;
      rank[i] = cost[i] = 0;
    }
  }
  void add(int k, int x) { cost[k] = x; }
  int find(int x) {
    if (x == par[x]) return x;
    return par[x] = find(par[x]);
  }
  int unite(int x, int y) {
    x = find(x);
    y = find(y);
    if (x == y) return 0;
    if (rank[x] < rank[y]) {
      par[x] = y;
      cost[y] += cost[x];
      return cost[y];
    } else {
      par[y] = x;
      cost[x] += cost[y];
      if (rank[x] == rank[y]) rank[x]++;
      return cost[x];
    }
  }
};
UF uf;
vector<pair<int, int> > vec[10005];
int main() {
  int n;
  scanf("%d", &n);
  if (n == 1) {
    puts("0");
    return 0;
  }
  for (int i = 0; i < n - 1; i++) {
    int a, b, c;
    scanf("%d %d %d", &a, &b, &c);
    a--;
    b--;
    vec[c].push_back(pair<int, int>(a, b));
  }
  uf.init(n + 2);
  int all = 0;
  for (int i = 0; i < n; i++) {
    int a;
    scanf("%d", &a);
    all += a;
    uf.add(i, a + 1);
  }
  int ret = 1;
  for (int i = 1; i < 10005; i++) {
    int gt = 0;
    for (int j = 0; j < vec[i].size(); j++) {
      pair<int, int> p = vec[i][j];
      gt = max(gt, uf.unite(p.first, p.second));
    }
    if (gt > all) break;
    ret = i + 1;
  }
  printf("%d\n", ret);
  return 0;
}