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;
FILE *debug;
void setIO(string file = "sample") {}
void closeIO() {}
template <class T>
void read(T &x) {
  char tmp = 0;
  T flag = 1, ret = 0;
  while (tmp != '-' && !isdigit(tmp)) tmp = getchar();
  if (tmp == '-') flag = -1, tmp = getchar();
  while (isdigit(tmp)) ret *= 10, ret += tmp - '0', tmp = getchar();
  x = flag * ret;
}
long long getint() {
  long long ret;
  read<long long>(ret);
  return ret;
}
void endline() { fprintf(debug, "\n"); }
pair<int, pair<int, int> > v[3010];
int n, times[3010], x[3010], fa[3010], xsum;
int gf(int x) {
  if (x == fa[x]) return x;
  return gf(fa[x]);
}
int main(int argc, char *argv[]) {
  setIO("sample");
  int i = 1;
  n = getint();
  for (i = 1; i <= n - 1; ++i)
    v[i].second.first = getint(), v[i].second.second = getint(),
    v[i].first = getint();
  sort(v + 1, v + n);
  for (i = 1; i <= n; fa[i] = i, ++i) {
    times[i] = x[i] = getint();
    xsum += x[i];
    ++times[i];
  }
  for (i = 1; i <= n - 1; ++i) {
    times[gf(v[i].second.second)] += times[gf(v[i].second.first)];
    fa[gf(v[i].second.first)] = fa[gf(v[i].second.second)];
    if (times[gf(v[i].second.second)] > xsum) break;
  }
  printf("%d", v[i].first);
  closeIO();
  return EXIT_SUCCESS;
}