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 std::cin;
using std::cout;
const int maxn = 10100;
int n, ans;
int sum = 0;
struct edge {
  int u, v, w;
} a[maxn];
int fa[maxn], size[maxn], can[maxn];
inline int find(int x) { return fa[x] == x ? fa[x] : fa[x] = find(fa[x]); }
inline int cmp0(const edge& x, const edge& y) { return x.w < y.w; }
inline int check(int x) { return size[find(x)] > sum - can[find(x)]; }
int main() {
  std::ios::sync_with_stdio(false), cin.tie(0);
  cin >> n;
  for (int i = 1; i < n; ++i) {
    cin >> a[i].u >> a[i].v >> a[i].w;
  }
  for (int i = 1; i <= n; ++i) {
    cin >> can[i];
    sum += can[i];
    fa[i] = i;
    size[i] = 1;
  }
  std::sort(a + 1, a + n, cmp0);
  for (int i = 1; i <= n; ++i) {
    ans += check(i);
  }
  for (int i = 1; i < n; ++i) {
    int x = find(a[i].u), y = find(a[i].v);
    ans -= check(x) + check(y);
    fa[x] = y;
    size[y] += size[x];
    can[y] += can[x];
    ans += check(x);
    if (ans) {
      cout << a[i].w << '\n';
      return 0;
    }
  }
  cout << a[n - 1].w << '\n';
}