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;
int a, b, c, d, e, f, g, h, i, j, k;
struct edge {
  int u, v, w;
  friend bool operator<(const edge &x, const edge &y) { return x.w < y.w; }
};
edge E[N];
int fa[N], size[N], D[N], ALL;
int Find(int x) { return fa[x] == x ? x : fa[x] = Find(fa[x]); }
void Init() {
  cin >> a;
  for (int ii = 1; ii < a; ii++) scanf("%d%d%d", &E[ii].u, &E[ii].v, &E[ii].w);
  sort(E + 1, E + a);
  for (int ii = 1; ii <= a; ii++) scanf("%d", &D[ii]), ALL += D[ii];
}
void Work() {
  bool ch = true;
  int an = 0;
  for (int ii = 1; ii <= a; ii++) fa[ii] = ii, size[ii] = 1;
  for (int ii = 1; ii <= a; ii++)
    if (size[ii] > ALL - D[ii]) ch = false;
  if (ch) an = E[1].w;
  for (int ii = 1; ch && ii < a - 1; ii++) {
    c = Find(E[ii].u), d = Find(E[ii].v);
    fa[d] = c, size[c] += size[d], D[c] += D[d];
    if (size[c] > ALL - D[c]) ch = false;
    if (ch) an = E[ii + 1].w;
  }
  cout << an << endl;
}
int main() {
  Init(), Work();
  return 0;
}