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 E {
  int u, v, w;
} e[3005];
int f[3005], size[3005], val[3005], sum, n;
bool cmp(E x, E y) { return x.w < y.w; }
int find(int x) { return f[x] == x ? x : f[x] = find(f[x]); }
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) scanf("%d%d%d", &e[i].u, &e[i].v, &e[i].w);
  for (int i = 1; i <= n; i++)
    scanf("%d", val + i), sum += val[i], size[i] = 1, f[i] = i;
  sort(e + 1, e + n, cmp);
  for (int i = 1; i < n; i++) {
    int x = e[i].u, y = e[i].v, fx = find(x), fy = find(y);
    f[fx] = fy, size[fy] += size[fx], val[fy] += val[fx];
    if (size[fy] > sum - val[fy]) return printf("%d", e[i].w), 0;
  }
  return printf("%d", e[n - 1].w), 0;
}