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 = 1000005;
const double eps = 1e-8;
const double DINF = 1e100;
const int INF = 1000000006;
const long long LINF = 1000000000000000005ll;
struct Edge {
  int a, b, c;
  Edge() {}
  Edge(int a, int b, int c) : a(a), b(b), c(c) {}
} edge[MaxN];
int fa[MaxN];
int n;
int get(int x) {
  if (x == fa[x]) return x;
  return fa[x] = get(fa[x]);
}
int x[MaxN];
int size[MaxN];
int sumx[MaxN];
int all = 0;
bool check(int w) {
  for (int i = 1; i <= n; ++i) {
    fa[i] = i;
    size[i] = 1;
    sumx[i] = x[i];
  }
  for (int i = 1; i < n; ++i) {
    int u = edge[i].a;
    int v = edge[i].b;
    int c = edge[i].c;
    if (c >= w) continue;
    int a = get(u);
    int b = get(v);
    fa[a] = b;
    size[b] += size[a];
    sumx[b] += sumx[a];
  }
  for (int i = 1; i <= n; ++i)
    if (fa[i] == i) {
      if (size[i] > all - sumx[i]) return false;
    }
  return true;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; ++i) {
    int a, b, c;
    scanf("%d %d %d", &a, &b, &c);
    edge[i] = Edge(a, b, c);
  }
  for (int i = 1; i <= n; ++i) {
    scanf("%d", x + i);
    all += x[i];
  }
  int l = 0, r = INF;
  while (l + 1 < r) {
    int mid = (l + r) >> 1;
    if (check(mid))
      l = mid;
    else
      r = mid;
  }
  printf("%d\n", l);
  return 0;
}