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;
const int inf = 1e9;
template <typename T>
void read(T &x) {
  x = 0;
  int f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) x = (x << 1) + (x << 3) + c - '0';
  x *= f;
}
int n;
int fa[N], siz[N], val[N];
int sum;
struct misaka {
  int x, y, w;
  bool operator<(const misaka &cmp) const { return w < cmp.w; }
} e[N];
int find(int x) {
  if (fa[x] == x) return x;
  return fa[x] = find(fa[x]);
}
int main() {
  read(n);
  for (int i = 1; i <= n - 1; i++) read(e[i].x), read(e[i].y), read(e[i].w);
  for (int i = 1; i <= n; i++)
    fa[i] = i, siz[i] = 1, read(val[i]), sum += val[i];
  sort(e + 1, e + n);
  for (int i = 1; i <= n - 1; i++) {
    int x = e[i].x, y = e[i].y;
    int fx = find(x), fy = find(y);
    siz[fx] += siz[fy], val[fx] += val[fy];
    fa[fy] = fx;
    if (siz[fx] > sum - val[fx]) return printf("%d\n", e[i].w), 0;
  }
  printf("%d\n", e[n - 1].w);
  return 0;
}