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>
const int oo = 0x3f3f3f3f;
template <typename T>
inline bool chkmax(T &a, T b) {
  return a < b ? a = b, true : false;
}
template <typename T>
inline bool chkmin(T &a, T b) {
  return a > b ? a = b, true : false;
}
template <typename T>
T read(T &x) {
  int f = 1;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') f = -1;
  for (x = 0; isdigit(ch); ch = getchar()) x = 10 * x + ch - '0';
  return x *= f;
}
const int MAXN = 3005;
struct Edge {
  int u, v;
  int w;
  bool operator<(const Edge &rhs) const { return w < rhs.w; }
};
int N, M;
int first[MAXN];
Edge E[MAXN];
int fa[MAXN];
int size[MAXN];
int sumx[MAXN];
int glbsumx;
int getFather(int u) { return fa[u] == u ? u : fa[u] = getFather(fa[u]); }
void input() {
  read(N);
  M = N - 1;
  for (int i = 0; i < M; ++i) {
    read(E[i].u);
    read(E[i].v);
    read(E[i].w);
  }
  for (int i = 1; i <= N; ++i) {
    read(first[i]);
    glbsumx += first[i];
  }
}
void solve() {
  memcpy(sumx + 1, first + 1, N * sizeof(*sumx));
  for (int i = 1; i <= N; ++i) {
    fa[i] = i;
    size[i] = 1;
  }
  std::sort(E, E + M);
  int i;
  for (i = 0; i < M; ++i) {
    int fu = getFather(E[i].u);
    int fv = getFather(E[i].v);
    assert(fu != fv);
    fa[fu] = fv;
    size[fv] += size[fu];
    sumx[fv] += sumx[fu];
    if (size[fv] > glbsumx - sumx[fv]) break;
  }
  printf("%d\n", E[i].w);
}
int main() {
  input();
  solve();
  return 0;
}