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 = 1e5 + 5;
int n;
struct edge {
  int u, v, w;
} e[N];
int fa[N], siz[N], x[N], s;
inline int findf(int x) { return fa[x] == x ? x : fa[x] = findf(fa[x]); }
inline bool cmp(const edge& a, const edge& b) { return a.w < b.w; }
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", &x[i]), fa[i] = i, siz[i] = 1, s += x[i];
  sort(e + 1, e + n, cmp);
  for (int i = 1; i < n; i++) {
    int fx = findf(e[i].u), fy = findf(e[i].v);
    if (fx == fy) continue;
    fa[fx] = fy, siz[fy] += siz[fx], x[fy] += x[fx];
    if (siz[fy] > s - x[fy]) {
      printf("%d", e[i].w);
      return 0;
    }
  }
  printf("%d", e[n - 1].w);
  return 0;
}