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 MAXN = 1e6 + 100;
int p[MAXN];
int root(int a) {
  if (p[a] < 0) return a;
  return p[a] = root(p[a]);
}
void merg(int a, int b) {
  a = root(a), b = root(b);
  p[a] += p[b];
  p[b] = a;
}
int d1[MAXN], d2[MAXN];
bool calc(int n) {
  int sum = 0;
  for (int i = 0; i < n; i++) sum += d2[i];
  if (sum < n) return false;
  for (int i = 0; i < n; i++)
    if (d1[i] > sum - d2[i]) return false;
  return true;
}
int a[MAXN];
vector<pair<pair<int, int>, int>> e;
bool check(int k, int n) {
  fill(p, p + n, -1);
  for (auto edge : e) {
    int x = edge.first.first, y = edge.first.second;
    int w = edge.second;
    if (w < k) merg(x, y);
  }
  fill(d2, d2 + n, 0);
  for (int i = 0; i < n; i++) {
    d1[i] = max(0, -p[i]);
    d2[root(i)] += a[i];
  }
  return calc(n);
}
int32_t main() {
  int n;
  cin >> n;
  for (int i = 1; i < n; i++) {
    int x, y, z;
    cin >> x >> y >> z;
    x--;
    y--;
    e.push_back({{x, y}, z});
  }
  for (int i = 0; i < n; i++) cin >> a[i], d2[i] = a[i];
  fill(d1, d1 + n, 1);
  if (!calc(n)) return cout << "0\n", 0;
  int l = 1;
  int r = 1e9;
  while (r - l > 1) {
    int mid = (r + l) / 2;
    if (check(mid, n))
      l = mid;
    else
      r = mid;
  }
  cout << l;
}