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 MAXN = 5000;
bool mark[MAXN];
int m[MAXN], ind, t[MAXN], ans, r, n, a[MAXN], sum, f[MAXN];
unsigned int l;
vector<int> G[MAXN], W[MAXN], v1;
void dfs(int v, int k) {
  mark[v] = true;
  m[v] = ind;
  t[ind]++;
  for (int i = 0; i < G[v].size(); i++) {
    int u = G[v][i];
    if (!mark[u] && W[v][i] <= k) dfs(u, k);
  }
}
bool check(int k) {
  memset(mark, false, sizeof mark);
  ind = 0;
  for (int i = 1; i <= n; i++)
    if (!mark[i]) {
      t[ind] = 0;
      f[ind] = 0;
      dfs(i, k);
      ind++;
    }
  ans = 0;
  for (int i = 1; i <= n; i++) f[m[i]] += a[i];
  for (int i = 0; i < ind; i++)
    if (sum - f[i] < t[i]) return false;
  for (int i = 1; i <= n; i++) ans += max(0, min(a[i], n - t[m[i]]));
  if (ans >= n) return true;
  return false;
}
int main() {
  cin >> n;
  for (int i = 1; i < n; i++) {
    int v, u, w;
    cin >> v >> u >> w;
    v1.push_back(w);
    G[v].push_back(u);
    G[u].push_back(v);
    W[v].push_back(w);
    W[u].push_back(w);
  }
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    sum += a[i];
  }
  if (n == 1) {
    cout << 0 << endl;
    return 0;
  }
  sort(v1.begin(), v1.end());
  r = v1.size() - 1;
  while (l < r) {
    int mid = (l + r + 1) / 2;
    if (check(v1[mid]))
      l = mid;
    else
      r = mid - 1;
  }
  if (l + 1 >= v1.size())
    cout << v1.back() << endl;
  else
    cout << v1[l + 1] << endl;
}