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 n, x[N], f[N], sz[N], v[N];
inline int getf(int v) { return f[v] == v ? v : f[v] = getf(f[v]); }
struct Edge {
  int u, v, w;
} e[N];
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i < n; i++) cin >> e[i].u >> e[i].v >> e[i].w;
  sort(e + 1, e + n, [](Edge a, Edge b) { return a.w < b.w; });
  int sum = 0;
  for (int i = 1; i <= n; i++) cin >> x[i], sum += x[i];
  for (int i = 1; i <= n; i++) f[i] = i, sz[i] = 1, v[i] = x[i];
  for (int i = 1; i < n; i++) {
    int fx = getf(e[i].u), fy = getf(e[i].v);
    f[fx] = fy;
    sz[fy] += sz[fx];
    v[fy] += v[fx];
    if (sz[fy] > sum - v[fy]) {
      cout << e[i].w << endl;
      return 0;
    }
  }
  cout << e[n - 1].w << endl;
  return 0;
}