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 long double eps = 1e-9;
const int inf = (1 << 30) - 1;
const long long inf64 = ((long long)1 << 62) - 1;
const long double pi = acos(-1);
template <class T>
T sqr(T x) {
  return x * x;
}
template <class T>
T abs(T x) {
  return x < 0 ? -x : x;
}
const int MAXN = 10 * 1000;
int n, sum_x;
pair<int, pair<int, int> > edg[MAXN];
int x[MAXN], color[MAXN], num[MAXN], sum[MAXN];
bool check(int w) {
  for (int i = 0; i < n; ++i) {
    color[i] = i;
  }
  for (int i = 0; i < n - 1 && edg[i].first < w; ++i) {
    int v1 = edg[i].second.first;
    int v2 = edg[i].second.second;
    if (color[v1] != color[v2]) {
      int c1 = color[v1];
      for (int i = 0; i < n; ++i) {
        if (color[i] == c1) {
          color[i] = color[v2];
        }
      }
    }
  }
  for (int i = 0; i < n; ++i) {
    num[i] = 0;
    sum[i] = 0;
  }
  for (int i = 0; i < n; ++i) {
    num[color[i]]++;
    sum[color[i]] += x[i];
  }
  for (int i = 0; i < n; ++i) {
    if (num[i] > sum_x - sum[i]) {
      return false;
    }
  }
  return true;
}
int main() {
  cin >> n;
  int max_w = 0;
  for (int i = 0; i < n - 1; ++i) {
    scanf("%d %d %d", &edg[i].second.first, &edg[i].second.second,
          &edg[i].first);
    edg[i].second.first--;
    edg[i].second.second--;
    max_w = max(max_w, edg[i].first);
  }
  sum_x = 0;
  for (int i = 0; i < n; ++i) {
    scanf("%d", &x[i]);
    sum_x += x[i];
  }
  sort(edg, edg + n - 1);
  int l = 0;
  int r = max_w;
  while (l < r) {
    int m = (l + r + 1) / 2;
    if (check(m)) {
      l = m;
    } else {
      r = m - 1;
    }
  }
  cout << l << endl;
  return 0;
}