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;
inline int read() {
  int s = 0, w = 1;
  char c;
  while (c = getchar(), c > '9' || c < '0')
    if (c == '-') w = -1;
  while (c >= '0' && c <= '9') s = s * 10 + c - '0', c = getchar();
  return s * w;
}
const int N = 1e5 + 7;
struct edge {
  int u, t, w;
  bool operator<(edge a) const { return w < a.w; }
} e[N];
int n, sum, ans, can;
int X[N], siz[N], fa[N];
int find(int x) { return x == fa[x] ? x : fa[x] = find(fa[x]); }
vector<pair<int, int> > V;
void merge(int a, int b) {
  fa[b] = a;
  siz[a] += siz[b];
  X[a] += X[b];
  if (siz[a] > sum - X[a]) can = 1;
}
int main() {
  n = read();
  if (n == 1) return puts("0"), 0;
  for (int i = 1; i < n; i++) e[i].u = read(), e[i].t = read(), e[i].w = read();
  sort(e + 1, e + n);
  for (int i = 1; i <= n; i++) X[i] = read(), sum += X[i];
  for (int i = 1; i <= n; i++) fa[i] = i, siz[i] = 1;
  for (int i = 1; i < n; i++) {
    if (can) break;
    ans = e[i].w;
    int u = find(e[i].u), t = find(e[i].t);
    merge(u, t);
  }
  printf("%d\n", ans);
}