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 N = 3008;
int n, fa[N], sz[N], sum[N], all, ans;
struct Edge {
  int u, v, d;
  bool operator<(const Edge &b) const { return d < b.d; }
} L[N];
int find(int x) { return x == fa[x] ? fa[x] : fa[x] = find(fa[x]); }
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n - 1; i++) {
    scanf("%d%d%d", &L[i].u, &L[i].v, &L[i].d);
  }
  for (int i = 1; i <= n; i++)
    scanf("%d", &sum[i]), fa[i] = i, sz[i] = 1, all += sum[i];
  sort(L + 1, L + n);
  ans = L[1].d;
  for (int i = 1; i <= n - 1; i++) {
    int u = find(L[i].u), v = find(L[i].v);
    sz[u] += sz[v];
    sum[u] += sum[v];
    fa[v] = u;
    if (sz[u] + sum[u] > all) break;
    if (L[i].d != L[i + 1].d) ans = L[i + 1].d;
  }
  printf("%d\n", ans);
  return 0;
}