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 edge {
  int a, b, v;
} e[3010];
int fa[3010], siz[3010], cnt[3010], sum;
bool can;
bool operator<(edge a, edge b) { return a.v < b.v; }
int getfa(int x) {
  if (fa[x] == x) return x;
  return fa[x] = getfa(fa[x]);
}
void merge(int x, int y) {
  x = getfa(x);
  y = getfa(y);
  siz[x] += siz[y];
  cnt[x] += cnt[y];
  fa[y] = x;
  if (siz[x] > sum - cnt[x]) can = 0;
}
int main() {
  int n, i, ans;
  scanf("%d", &n);
  for (i = 1; i < n; i++) scanf("%d%d%d", &e[i].a, &e[i].b, &e[i].v);
  for (i = 1; i <= n; i++) {
    scanf("%d", cnt + i);
    sum += cnt[i];
    siz[i] = 1;
    fa[i] = i;
  }
  sort(e + 1, e + n);
  ans = 0;
  can = 1;
  for (i = 1; i < n; i++) {
    if (can) ans = e[i].v;
    merge(e[i].a, e[i].b);
  }
  printf("%d", ans);
}