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;
constexpr int MAXN = 3e3 + 4;
constexpr int MAXA = 2e3 + 5;
constexpr int MOD = 1e9 + 7;
constexpr long long INF = 1e18;
constexpr int LOG = 40;
int n, a[MAXN], b[MAXN], c[MAXN], x[MAXN], par[MAXN], s[MAXN], mx, idx;
vector<int> vc;
int get_par(int u) { return (par[u] < 0 ? u : par[u] = get_par(par[u])); }
void merge(int u, int v) {
  if ((u = get_par(u)) == (v = get_par(v))) {
    return;
  }
  if (par[u] > par[v]) {
    swap(u, v);
  }
  par[u] += par[v];
  s[u] += s[v];
  par[v] = u;
  if (-par[u] > mx) {
    mx = -par[u];
    idx = u;
  }
}
bool Check(int y) {
  memset(par, -1, sizeof par);
  for (int i = 1; i <= n; i++) {
    s[i] = x[i];
  }
  mx = 1, idx = 1;
  for (int i = 1; i <= n; i++) {
    if (c[i] < y) {
      merge(a[i], b[i]);
    }
  }
  if (mx <= n / 2) {
    return true;
  }
  int sum = 0;
  for (int i = 1; i <= n; i++) {
    if (par[i] < 0 && i != idx) {
      sum += s[i] + par[i];
    }
  }
  int tmp = mx - (n - mx);
  if (tmp <= sum) {
    return true;
  }
  return false;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i < n; i++) {
    cin >> a[i] >> b[i] >> c[i];
    vc.push_back(c[i]);
  }
  for (int i = 1; i <= n; i++) {
    cin >> x[i];
  }
  sort(vc.begin(), vc.end());
  int l = 0, r = 0;
  if (!vc.empty()) {
    l = vc[0], r = vc.back() + 1;
  }
  while (l + 1 < r) {
    int mid = (l + r) >> 1;
    if (Check(mid)) {
      l = mid;
    } else {
      r = mid;
    }
  }
  cout << l;
  return 0;
}