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 = 2e6 + 100;
struct E {
  int u, v, w;
} e[N * 2];
int siz[N], sum[N], fa[N];
int x[N], xi, book[N], n, ans;
int L, R;
bool C1(E a, E b) { return a.w < b.w; }
int find(int x) {
  if (fa[x] == x) return x;
  return fa[x] = find(fa[x]);
}
bool check(int X) {
  for (int i = 1; i <= n; i++)
    fa[i] = i, siz[i] = 1, sum[i] = x[i], book[i] = 0;
  for (int i = 1; i <= n - 1; i++) {
    if (e[i].w >= X) break;
    int dx = find(e[i].u), dy = find(e[i].v);
    siz[dy] += siz[dx];
    siz[dx] = 0;
    sum[dy] += sum[dx];
    sum[dx] = 0;
    fa[dx] = dy;
  }
  for (int i = 1; i <= n; i++) {
    int dx = find(i);
    if (book[dx]) continue;
    book[dx] = 1;
    if (siz[dx] > xi - sum[dx]) return false;
  }
  return true;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n - 1; i++)
    scanf("%d%d%d", &e[i].u, &e[i].v, &e[i].w), R = max(R, e[i].w);
  for (int i = 1; i <= n; i++) scanf("%d", &x[i]), xi += x[i];
  sort(e + 1, e + n, C1);
  while (L <= R) {
    int mid = (L + R) >> 1;
    if (check(mid))
      ans = mid, L = mid + 1;
    else
      R = mid - 1;
  }
  printf("%d\n", ans);
  return 0;
}