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 N = 3005;
int n, ans, sum, a[N];
int sz[N], fa[N];
struct edge {
  int x, y, v;
  bool operator<(const edge &a) const { return v < a.v; }
} e[N];
int get(int x) { return x == fa[x] ? x : fa[x] = get(fa[x]); }
int main() {
  scanf("%d", &n);
  for (int i = (int)(1); i <= (int)(n - 1); i++)
    scanf("%d%d%d", &e[i].x, &e[i].y, &e[i].v);
  sort(e + 1, e + n);
  for (int i = (int)(1); i <= (int)(n); i++) scanf("%d", &a[i]);
  for (int i = (int)(1); i <= (int)(n); i++) sz[i] = 1, fa[i] = i, sum += a[i];
  for (int i = (int)(1); i <= (int)(n - 1); i++) {
    ans = e[i].v;
    int x = get(e[i].x), y = get(e[i].y);
    fa[x] = y;
    sz[y] += sz[x];
    a[y] += a[x];
    if (sz[y] > sum - a[y]) break;
  }
  printf("%d\n", ans);
}