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;
struct node {
  int x, y, z;
} e[5010];
bool cmp(const node& a, const node& b) { return a.z < b.z; }
int f[5010];
int find(int x) { return f[x] == x ? x : f[x] = find(f[x]); }
int n, bz[5010], sz[5010];
int main() {
  cin >> n;
  for (int i = 1; i < n; i++) cin >> e[i].x >> e[i].y >> e[i].z;
  int tot = 0;
  for (int i = 1; i <= n; i++) cin >> sz[i], f[i] = i, tot += sz[i], bz[i] = 1;
  sort(e + 1, e + n, cmp);
  int ans = e[n - 1].z;
  bool flag = true;
  for (int i = 1; i < n;) {
    int j = i;
    for (; i < n && e[i].z == e[j].z; i++) {
      int fx = find(e[i].x), fy = find(e[i].y);
      f[fy] = fx;
      sz[fx] += sz[fy];
      bz[fx] += bz[fy];
      if (bz[fx] > tot - sz[fx]) flag = false;
    }
    if (!flag) {
      ans = e[j].z;
      break;
    }
  }
  cout << ans << endl;
  return 0;
}