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;
typedef struct N {
  int have, need;
} N;
typedef struct E {
  int a, b, len;
} E;
typedef struct G {
  int par, rnk, have, need;
} G;
int n;
N node[3000];
E edge[2999];
G grp[3000];
int p[2999];
bool elencmp(const int &a, const int &b) { return edge[a].len < edge[b].len; }
int find(int a) {
  if (grp[a].par == a) return a;
  return grp[a].par = find(grp[a].par);
}
void run() {
  scanf("%d", &n);
  for (int i = (0); i < (n - 1); ++i) {
    int a, b, c;
    scanf("%d%d%d", &a, &b, &c);
    --a, --b;
    edge[i].a = a;
    edge[i].b = b;
    edge[i].len = c;
  }
  for (int i = (0); i < (n); ++i) node[i].need = 1, scanf("%d", &node[i].have);
  for (int i = (0); i < (n - 1); ++i) p[i] = i;
  sort(p, p + n - 1, elencmp);
  for (int i = (0); i < (n); ++i)
    grp[i].par = i, grp[i].rnk = 0, grp[i].have = node[i].have,
    grp[i].need = node[i].need;
  int sumhave = 0;
  for (int i = (0); i < (n); ++i) sumhave += node[i].have;
  int ret = 0;
  for (int i = (0); i < (n - 1); ++i) {
    ret = edge[p[i]].len;
    int a = find(edge[p[i]].a), b = find(edge[p[i]].b);
    if (a != b) {
      if (grp[a].need + grp[b].need > sumhave - grp[a].have - grp[b].have)
        break;
      if (grp[a].rnk < grp[b].rnk) swap(a, b);
      grp[a].have += grp[b].have, grp[a].need += grp[b].need,
          grp[b].par = a, grp[a].rnk += grp[a].rnk == grp[b].rnk ? 1 : 0;
    }
  }
  printf("%d\n", ret);
}
int main() {
  run();
  return 0;
}