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;
int N;
int x[3000];
int a[3000], b[3000], l[3000], cid[3000];
bool cmp(int i, int j) { return l[i] < l[j]; }
int fid[3000], sz[3000];
long long xsz[3000];
int imx;
void init() {
  for (int i = 0; i < N; i++) fid[i] = i, sz[i] = 1, xsz[i] = x[i];
}
int find(int i) {
  if (fid[i] == i) return i;
  fid[i] = find(fid[i]);
  return fid[i];
}
void join(int i, int j) {
  i = find(i), j = find(j);
  if (sz[i] <= sz[j]) {
    fid[i] = j;
    sz[j] += sz[i];
    xsz[j] += xsz[i];
    if (sz[j] > sz[imx]) imx = j;
  } else {
    fid[j] = i;
    sz[i] += sz[j];
    xsz[i] += xsz[j];
    if (sz[i] > sz[imx]) imx = i;
  }
}
int main() {
  cin >> N;
  for (int i = 0; i < N - 1; i++) {
    cin >> a[i] >> b[i] >> l[i];
    a[i]--, b[i]--;
    cid[i] = i;
  }
  long long tot = 0;
  for (int i = 0; i < N; i++) {
    cin >> x[i];
    tot += x[i];
  }
  sort(cid, cid + N - 1, cmp);
  init();
  for (int i = 0; i < N - 1; i++) {
    join(a[cid[i]], b[cid[i]]);
    if (sz[imx] + xsz[imx] > tot) {
      cout << l[cid[i]] << '\n';
      return 0;
    }
  }
  cout << 0 << '\n';
}