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 = 3005;
int n;
vector<pair<int, int> > edge[N];
int c[N];
int fa[N];
int cnt[N], sigma[N];
int find(int u) { return u == fa[u] ? u : fa[u] = find(fa[u]); }
void merge(int u, int v) { fa[v] = u; }
void dfs(int u, int bar, int fa = -1) {
  for (int i = 0; i < (int)edge[u].size(); ++i) {
    int v = edge[u][i].first, d = edge[u][i].second;
    if (v == fa) {
      continue;
    }
    dfs(v, bar, u);
    if (d < bar) {
      merge(find(v), find(u));
    }
  }
}
bool check(int bar) {
  for (int i = 0; i < n; ++i) {
    fa[i] = i;
  }
  dfs(0, bar);
  for (int i = 0; i < n; ++i) {
    cnt[i] = 0, sigma[i] = 0;
  }
  for (int i = 0; i < n; ++i) {
    ++cnt[find(i)], sigma[find(i)] += c[i];
  }
  int maxcnt = find(0), sum = 0;
  for (int i = 0; i < n; ++i) {
    if (find(i) == i) {
      if (cnt[i] > cnt[maxcnt]) {
        maxcnt = i;
      }
      sum += sigma[i];
    }
  }
  return sum - sigma[maxcnt] >= cnt[maxcnt];
}
int main() {
  scanf("%d", &n);
  if (n == 1) {
    printf("0\n");
    return 0;
  }
  vector<int> val;
  for (int i = 1; i < n; ++i) {
    int u, v, w;
    scanf("%d%d%d", &u, &v, &w);
    --u, --v;
    edge[u].push_back(make_pair(v, w));
    edge[v].push_back(make_pair(u, w));
    val.push_back(w);
  }
  sort(val.begin(), val.end());
  val.erase(unique(val.begin(), val.end()), val.end());
  for (int i = 0; i < n; ++i) {
    scanf("%d", c + i);
  }
  int l = 0, r = (int)val.size() - 1;
  while (l < r) {
    int m = l + r + 1 >> 1;
    if (check(val[m])) {
      l = m;
    } else {
      r = m - 1;
    }
  }
  printf("%d\n", val[l]);
  return 0;
}