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;
int n;
struct node {
  int x, y, v;
} d[5200];
inline bool cmp(node a, node b) { return a.v < b.v; }
int fa[5200], sz[5200], sum[5200], SUM;
inline int getroot(int x) {
  if (fa[x] == x) return x;
  return fa[x] = getroot(fa[x]);
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i < n; i++) {
    cin >> d[i].x >> d[i].y >> d[i].v;
  }
  for (int i = 1; i <= n; i++) cin >> sum[i], SUM += sum[i];
  sort(d + 1, d + n, cmp);
  for (int i = 1; i <= n; i++) fa[i] = i, sz[i] = 1;
  int last = 0;
  for (int i = 1; i < n; i++) {
    last = d[i].v;
    if (SUM - (sum[getroot(d[i].x)] + sum[getroot(d[i].y)]) <
        sz[getroot(d[i].x)] + sz[getroot(d[i].y)]) {
      ;
      break;
    } else
      sum[getroot(d[i].x)] += sum[getroot(d[i].y)],
          sz[getroot(d[i].x)] += sz[getroot(d[i].y)],
          fa[getroot(d[i].y)] = fa[getroot(d[i].x)];
  }
  cout << last << endl;
}