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 inf = ~0U >> 1;
const long long INF = (long long)1e18;
const double pi = acos(-1.0);
template <class T>
inline T sqr(T a) {
  return a * a;
}
template <class T>
inline T min(T a, T b, T c) {
  return min(min(a, b), c);
}
template <class T>
inline T max(T a, T b, T c) {
  return max(max(a, b), c);
}
template <class T>
inline void read(T &n) {
  char c;
  for (c = getchar(); !(c >= '0' && c <= '9'); c = getchar())
    ;
  n = c - '0';
  for (c = getchar(); c >= '0' && c <= '9'; c = getchar()) n = n * 10 + c - '0';
}
int pw(int base, int n, int mo) {
  if (n == 0) return 1;
  if (n == 1) return base;
  int tmp = pw(base, n >> 1, mo);
  tmp = (long long)tmp * tmp % mo;
  if (n & 1) tmp = (long long)tmp * base % mo;
  return tmp;
}
const int maxn = 3200;
int N, lim[maxn], f[maxn], x[maxn], y[maxn], w[maxn], sum[maxn], size[maxn], S;
vector<pair<int, int> > E[maxn];
int find(int x) { return (f[x] == x) ? x : (f[x] = find(f[x])); }
int check(int K) {
  for (int i = (1); i <= (N); ++i) f[i] = i, size[i] = sum[i] = 0;
  for (int i = (1); i <= (N - 1); ++i)
    if (w[i] < K) f[find(x[i])] = find(y[i]);
  for (int i = (1); i <= (N); ++i) f[i] = find(f[i]);
  for (int i = (1); i <= (N); ++i) size[f[i]]++, sum[f[i]] += lim[i];
  for (int i = (1); i <= (N); ++i)
    if (size[i] > S - sum[i]) return 0;
  return 1;
}
int main() {
  scanf("%d", &N);
  for (int i = (1); i <= (N - 1); ++i) scanf("%d%d%d", &x[i], &y[i], &w[i]);
  for (int i = (1); i <= (N); ++i) scanf("%d", &lim[i]), S += lim[i];
  int l = 0, r = 20000;
  while (l + 1 < r) {
    int m = l + r >> 1;
    if (check(m))
      l = m;
    else
      r = m;
  }
  while (!check(l) && l) l--;
  while (check(l + 1)) l++;
  printf("%d\n", l);
  return 0;
}