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 maxN = 3e3 + 5;
struct Edge {
  int u, v, w;
  bool operator<(const Edge& t) const { return w < t.w; }
} E[maxN];
int f[maxN];
int N;
int _sum;
int sum[maxN];
int sz[maxN];
int x[maxN];
int get_f(int x) { return f[x] == x ? x : f[x] = get_f(f[x]); }
int main() {
  scanf("%d", &N);
  for (int i = 1; i <= N - 1; i++) {
    scanf("%d%d%d", &E[i].u, &E[i].v, &E[i].w);
  }
  for (int i = 1; i <= N; i++) {
    scanf("%d", &x[i]);
    sum[i] = x[i];
    sz[i] = 1;
    f[i] = i;
    _sum += x[i];
  }
  sort(E + 1, E + N);
  int ans = 0;
  for (int i = 1; i < N; i++) {
    int u = E[i].u;
    int v = E[i].v;
    int fu = get_f(u);
    int fv = get_f(v);
    if (sz[fu] + sz[fv] > _sum - sum[fu] - sum[fv]) {
      ans = E[i].w;
      break;
    }
    f[fu] = fv;
    sz[fv] += sz[fu];
    sum[fv] += sum[fu];
  }
  printf("%d\n", ans);
  return 0;
}