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 = 5000 + 7;
const int M = 1007;
const int inf = 1e9 + 7;
const long long linf = 1ll * inf * (inf - 1);
const double pi = acos(-1);
const double eps = 1e-7;
const bool multipleTest = 0;
vector<pair<pair<int, int>, int> > edge;
int r[N], tol[N];
int limit[N];
int n;
int get(int u) { return r[u] < 0 ? u : r[u] = get(r[u]); }
void unite(int u, int v) {
  u = ::get(u);
  v = ::get(v);
  if (r[u] > r[v]) swap(u, v);
  r[u] += r[v];
  r[v] = u;
  limit[u] += limit[v];
  tol[u] += tol[v];
}
void solve() {
  cin >> n;
  if (n == 1) {
    puts("0");
    return;
  }
  for (int i = (1); i < (n); i++) {
    int u, v, c;
    scanf("%d%d%d", &u, &v, &c);
    edge.push_back(make_pair(make_pair(c, u), v));
  }
  int sum = 0;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", limit + i);
    r[i] = -1;
    sum += limit[i];
    tol[i] = 1;
  }
  sort((edge).begin(), (edge).end());
  int ans = 0;
  if (sum >= n && n > 1) ans = edge[0].first.first;
  int last = edge[0].second;
  for (int i = 0; i < edge.size(); ++i) {
    int u = edge[i].first.second, v = edge[i].second;
    int c = edge[i].first.first;
    int rr = get(last);
    if (sum - limit[rr] >= tol[rr] &&
        sum - limit[rr] - tol[rr] + limit[rr] >= n - tol[rr]) {
      ans = c;
    } else
      break;
    unite(u, v);
    last = v;
  }
  cout << ans << '\n';
}
void test() {
  freopen("in.txt", "w", stdout);
  for (int i = 0; i < 50000; ++i) {
    printf("%c", ((rand() % 26) + 'a'));
  }
  cout << '\n' << 100000 << '\n';
  for (int i = (0); i < (100000); i++) {
    for (int t = (0); t < ((rand() % 4) + 1); t++)
      printf("%c", ((rand() % 26) + 'a'));
    printf(" ");
    for (int t = (0); t < ((rand() % 4) + 1); t++)
      printf("%c", ((rand() % 26) + 'a'));
    printf("\n");
  }
}
int main() {
  int Test = 1;
  if (multipleTest) {
    cin >> Test;
  }
  for (int i = 0; i < Test; ++i) {
    solve();
  }
}