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 = 100010;
int n, fa[maxn], size[maxn], a[maxn], sum = 0;
struct node {
  int u, v, w;
} t[maxn];
bool cmp(node x, node y) { return x.w < y.w; }
int find(int x) {
  if (x == fa[x]) return x;
  return fa[x] = find(fa[x]);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) scanf("%d%d%d", &t[i].u, &t[i].v, &t[i].w);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    fa[i] = i;
    size[i] = 1;
    sum += a[i];
  }
  sort(t + 1, t + n, cmp);
  for (int i = 1; i < n; i++) {
    int fx = find(t[i].u), fy = find(t[i].v);
    if (fx == fy) continue;
    fa[fx] = fy;
    size[fy] += size[fx];
    a[fy] += a[fx];
    if (size[fy] > sum - a[fy]) {
      printf("%d\n", t[i].w);
      return 0;
    }
  }
  printf("%d\n", t[n - 1].w);
  return 0;
}