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;
int n, hd[3003], nt[6003], nd[6003], wi[6003], t, m[3003], cnt[2][3003],
    x[3003];
void add(int u, int v, int w) {
  nt[t] = hd[u];
  nd[t] = v;
  wi[t] = w;
  hd[u] = t++;
}
bool vis[3003];
void dfs(int u, int w, int num) {
  vis[u] = 1;
  ++cnt[0][num], cnt[1][num] += x[u];
  for (int i = hd[u]; ~i; i = nt[i]) {
    if (wi[i] >= w) continue;
    int v = nd[i];
    if (vis[v]) continue;
    dfs(v, w, num);
  }
}
bool ck(int u) {
  t = 0, memset(vis, 0, sizeof vis), memset(cnt, 0, sizeof cnt);
  for (int i = 1; i <= n; ++i)
    if (!vis[i]) dfs(i, u, ++t);
  int sum = 0;
  for (int i = 1; i <= t; ++i) sum += cnt[1][i];
  for (int i = 1; i <= t; ++i)
    if (cnt[0][i] > sum - cnt[1][i]) return 0;
  return 1;
}
int main() {
  memset(hd, -1, sizeof hd);
  scanf("%d", &n);
  for (int i = 1, u, v, w; i < n; ++i) {
    scanf("%d%d%d", &u, &v, &w);
    add(u, v, w), add(v, u, w), m[i] = w;
  }
  sort(m + 1, m + n);
  for (int i = 1; i <= n; ++i) scanf("%d", &x[i]);
  int l = 0, r = n, mid;
  while (l < r - 1) {
    mid = (l + r) / 2;
    if (ck(m[mid]))
      l = mid;
    else
      r = mid;
  }
  printf("%d", m[l]);
}