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 = 3030, M = 1e4 + 100;
int n, S, ans, u[N], v[N], w[N], par[N], sz[N], cnt[N];
bool bad;
vector<int> vec[M];
int g_par(int v) { return v == par[v] ? v : par[v] = g_par(par[v]); }
int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    cin >> u[i] >> v[i] >> w[i];
    u[i]--;
    v[i]--;
    vec[w[i]].push_back(i);
  }
  for (int i = 0; i < n; i++) {
    cin >> sz[i];
    par[i] = i;
    cnt[i] = 1;
    S += sz[i];
  }
  for (int w = 0; w < M && bad == false; w++) {
    ans = w;
    for (int dex : vec[w]) {
      int U = g_par(u[dex]), V = g_par(v[dex]);
      sz[U] += sz[V];
      cnt[U] += cnt[V];
      par[V] = U;
      if (S - sz[U] < cnt[U]) bad = true;
    }
  }
  cout << (n == 1 ? 0 : ans);
  return 0;
}