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 = 3e3 + 1;
struct YLS {
  int a, b, w;
};
vector<YLS> srt;
bool cmp(const YLS &a, const YLS &b) { return (a.w < b.w); }
int pr[N], sz[N], x[N], n, m, a, b, w, ans, sm;
int find(int a) { return (!pr[a] ? a : find(pr[a])); }
int main() {
  YLS yls;
  scanf("%d", &n);
  for (int i = 1; i < n; i++)
    scanf("%d%d%d", &yls.a, &yls.b, &yls.w), srt.push_back(yls);
  sort(srt.begin(), srt.end(), cmp);
  for (int i = 1; i <= n; i++) scanf("%d", x + i), sz[i] = 1, sm += x[i];
  for (auto y : srt) {
    a = find(y.a), b = find(y.b);
    pr[b] = a, sz[a] += sz[b], x[a] += x[b], ans = y.w;
    if (sm - x[a] < sz[a]) break;
  }
  printf("%d", ans);
}