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;
template <class T, class U>
inline void smin(T &a, U b) {
  if (a > b) a = b;
}
template <class T, class U>
inline void smax(T &a, U b) {
  if (a < b) a = b;
}
template <class T>
inline void gn(T &x) {
  char c, sg = 0;
  while (c = getchar(), (c > '9' || c < '0') && c != '-' && c != -1)
    ;
  if (c == -1) return;
  if (c == '-') sg = 1, c = getchar();
  for (x = 0; c >= '0' && c <= '9';
       x = (x << 1) + (x << 3) + c - '0', c = getchar())
    ;
  if (sg) x = -x;
}
template <class T>
void print(T x) {
  if (x < 0) return putchar('-'), print(-x);
  if (x < 10) {
    putchar(x + '0');
    return;
  }
  print(x / 10);
  putchar(x % 10 + '0');
}
template <class U, class V>
inline void gn(U &u, V &v) {
  gn(u);
  gn(v);
}
template <class U, class V, class W>
inline void gn(U &u, V &v, W &w) {
  gn(u);
  gn(v, w);
}
template <class T>
void println(T x) {
  print(x);
  putchar('\n');
}
template <class T>
void printsp(T x) {
  print(x);
  putchar(' ');
}
template <class U, class V>
inline void print(U u, V v) {
  printsp(u);
  println(v);
}
template <class U, class V, class W>
inline void print(U u, V v, W w) {
  printsp(u);
  print(v, w);
}
int power(int a, int b, int m, int ans = 1) {
  for (; b; a = (long long)a * a % m, b >>= 1)
    if (b & 1) ans = (long long)ans * a % m;
  return ans;
}
pair<pair<int, int>, int> edge[3333];
int a[3333], tmp[3333], root[3333], sz[3333];
void init(int n) {
  for (int i = 1; i <= n; i++) root[i] = i, sz[i] = 1, tmp[i] = a[i];
}
int find(int first) {
  while (first != root[first]) first = root[first];
  return first;
}
void unite(int a, int b) {
  a = find(a);
  b = find(b);
  if (a == b) return;
  if (sz[a] < sz[b]) swap(a, b);
  sz[a] += sz[b];
  root[b] = a;
  tmp[a] += tmp[b];
}
int main() {
  int n;
  gn(n);
  for (int i = 0; i < n - 1; i++) {
    gn(edge[i].first.first, edge[i].first.second, edge[i].second);
  }
  int sum = 0;
  for (int i = 1; i <= n; i++) {
    gn(a[i]);
    sum += a[i];
  }
  int st = 0, ed = 10001;
  while (ed - st > 1) {
    int md = (ed + st) / 2;
    init(n);
    for (int i = 0; i < n - 1; i++) {
      if (edge[i].second < md) {
        unite(edge[i].first.first, edge[i].first.second);
      }
    }
    int vst = 0;
    for (int i = 1; i <= n; i++)
      if (i == find(i)) {
        if (sz[i] > sum - tmp[i]) {
          vst = 1;
          break;
        }
      }
    if (vst)
      ed = md;
    else
      st = md;
  }
  println(st);
  return 0;
}