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;
struct arr {
  int x, y, z;
} a[6005];
int n, sum = 0, mx[3005], fa[3005], sz[3005];
inline bool cmp(arr A, arr B) { return A.z < B.z; }
inline int find(int x) { return (fa[x] == x) ? x : fa[x] = find(fa[x]); }
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) fa[i] = i, sz[i] = 1;
  int x, y, z;
  for (int i = 1; i < n; i++) scanf("%d%d%d", &a[i].x, &a[i].y, &a[i].z);
  for (int i = 1; i <= n; i++) scanf("%d", &mx[i]), sum += mx[i];
  sort(a + 1, a + n, cmp);
  for (int i = 1; i < n; i++) {
    int fx = find(a[i].x);
    int fy = find(a[i].y);
    if (fx == fy) continue;
    fa[fy] = fx;
    sz[fx] += sz[fy];
    mx[fx] += mx[fy];
    if (sz[fx] > sum - mx[fx]) {
      printf("%d\n", a[i].z);
      return 0;
    }
  }
  printf("%d\n", a[n - 1].z);
  return 0;
}