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 node {
  int x, y, num;
};
node e[3001];
int n;
int ans = 0;
int father[3001];
int s[3001];
int outlet[3001];
int sum_out = 0;
bool cmp1(node xx, node yy) { return xx.num < yy.num; }
int find(int x) {
  if (father[x] != x) {
    father[x] = find(father[x]);
  }
  return father[x];
}
void unionset(int x, int y) {
  father[x] = y;
  s[y] += s[x];
  outlet[y] += outlet[x];
  return;
}
int main() {
  int i, j, k;
  int xx, yy;
  cin >> n;
  for (i = 1; i <= n - 1; i++) {
    cin >> e[i].x >> e[i].y >> e[i].num;
  }
  sort(e + 1, e + n, cmp1);
  for (i = 1; i <= n; i++) {
    father[i] = i;
    s[i] = 1;
    cin >> outlet[i];
    sum_out += outlet[i];
  }
  if (n == 1) {
    cout << 0 << endl;
    return 0;
  }
  ans = e[1].num;
  for (i = 1; i <= n - 2; i++) {
    xx = find(e[i].x);
    yy = find(e[i].y);
    unionset(xx, yy);
    if (sum_out - outlet[yy] < s[yy]) {
      break;
    } else {
      ans = e[i + 1].num;
    }
  }
  cout << ans << endl;
  return 0;
}