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 = (int)1E9;
int N, pro[3005], rep[3005], rnk[3005], rem[3005];
pair<int, pair<int, int> > es[3005];
int fd(int x) { return rep[x] == x ? x : rep[x] = fd(rep[x]); }
void us(int a, int b) {
  int x = fd(a), y = fd(b);
  if (x == y) return;
  if (rnk[x] > rnk[y]) swap(x, y);
  if (rnk[x] == rnk[y]) rnk[y]++;
  rep[x] = y;
  rem[y] += rem[x];
  pro[y] += pro[x];
}
int main() {
  cin >> N;
  if (N == 1) return puts("0"), 0;
  for (int i = (0); i < (N - 1); i++) {
    int a, b, c;
    scanf("%d%d%d", &a, &b, &c);
    a--;
    b--;
    es[i] = make_pair(c, pair<int, int>(a, b));
  }
  sort(es, es + N - 1);
  int sumpro = 0;
  for (int i = (0); i < (N); i++) {
    rem[i] = 1;
    rnk[i] = 0;
    rep[i] = i;
    scanf("%d", pro + i);
    sumpro += pro[i];
  }
  int tot = N, ans = es[0].first;
  for (int i = (0); i < (N - 1); i++) {
    int a = es[i].second.first, b = es[i].second.second;
    int x = fd(a), y = fd(b);
    us(x, y);
    int z = fd(x);
    if (rem[z] > sumpro - pro[z]) break;
    if (i + 1 < N - 1) ans = es[i + 1].first;
  }
  printf("%d\n", ans);
  return 0;
}