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 = 1.1e6;
int s[N], t[N], w[N], p[N], n, m, a[N], fa[N], sz[N], sum;
int cmp(int x, int y) { return w[x] < w[y]; }
int fd(int x) { return x == fa[x] ? x : fa[x] = fd(fa[x]); }
int main() {
  cin >> n;
  m = n - 1;
  for (int i = 1, x, y, z; i <= m; i++) {
    scanf("%d%d%d", &x, &y, &z);
    s[i] = x, t[i] = y, w[i] = z;
    p[i] = i;
  }
  sort(p + 1, p + m + 1, cmp);
  int sum = 0;
  for (int i = 1; i <= n; i++)
    scanf("%d", &a[i]), fa[i] = i, sz[i] = 1, sum += a[i];
  for (int i = 1, x, y, z; i <= m; i++) {
    x = s[p[i]], y = t[p[i]], z = w[p[i]];
    if (rand() % 2) swap(x, y);
    x = fd(x), y = fd(y);
    sz[y] += sz[x];
    a[y] += a[x];
    fa[x] = y;
    if (sz[y] > sum - a[y] || i == m) return cout << z, 0;
  }
  puts("0");
}