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 i64 = long long;
using namespace std;
const int nxM = int(1e4);
struct edge {
  int u, v, w;
};
bool cmp(const edge &a, const edge &b) { return a.w < b.w; }
i64 n, sz[nxM], sum_x[nxM], par[nxM], s, x[nxM];
vector<edge> adj;
int find(int u) {
  if (par[u] != u) par[u] = find(par[u]);
  return par[u];
}
void join(int u, int v) {
  u = find(u);
  v = find(v);
  if (u != v) {
    sz[u] += sz[v];
    sum_x[u] += sum_x[v];
    par[v] = u;
  }
}
void reset() {
  for (int i = 1; i <= n; ++i) {
    par[i] = i;
    sz[i] = 1;
    sum_x[i] = x[i];
  }
}
void input() {
  cin >> n;
  for (int i = 0; i < n - 1; ++i) {
    int a, b, c;
    cin >> a >> b >> c;
    adj.push_back({a, b, c});
  }
  for (int i = 1; i <= n; ++i) {
    cin >> x[i];
    s += x[i];
  }
}
bool solve(int m) {
  reset();
  for (edge &e : adj) {
    if (e.w < m) join(e.u, e.v);
  }
  for (int i = 1; i <= n; ++i) {
    if (find(i) == i) {
      if (sz[i] > s - sum_x[i]) return false;
    }
  }
  return true;
}
void binary(int l, int r) {
  while (true) {
    int m = l + (r - l) / 2;
    if (l == m || r == m) break;
    if (solve(m))
      l = m;
    else
      r = m;
  }
  for (int i = r; i >= l; --i) {
    if (solve(i)) {
      cout << i << endl;
      return;
    }
  }
  cout << "0" << endl;
}
int main() {
  input();
  binary(1, int(1e4));
  return 0;
}