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;
inline long long read() {
  long long x = 0, f = 1, ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
inline void write(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
inline void writeln(long long x) {
  write(x);
  puts("");
}
inline void write_p(long long x) {
  write(x);
  putchar(' ');
}
const int N = 200005;
int n, xx[N], sz[N], Fa[N], vis[N], cnt, poi[N], nxt[N], head[N], top;
struct node {
  int x, y, v;
} e[N];
inline bool cmp(node x, node y) { return x.v < y.v; }
pair<int, int> q[N];
inline void add(int x, int y) {
  poi[++cnt] = y;
  nxt[cnt] = head[x];
  head[x] = cnt;
  poi[++cnt] = x;
  nxt[cnt] = head[y];
  head[y] = cnt;
}
inline pair<int, int> Dfs(int x, int fa) {
  vis[x] = 1;
  pair<int, int> tmp = make_pair(1, xx[x]);
  for (int i = head[x]; i; i = nxt[i]) {
    if (poi[i] == fa) continue;
    pair<int, int> las = Dfs(poi[i], x);
    tmp.first += las.first;
    tmp.second += las.second;
  }
  return tmp;
}
inline bool check(int line) {
  for (int i = 1; i <= n; ++i) head[i] = 0;
  cnt = 0;
  for (int i = 1; i <= n - 1; ++i)
    if (e[i].v < line)
      add(e[i].x, e[i].y);
    else
      break;
  for (int i = 1; i <= n; ++i) vis[i] = 0;
  top = 0;
  int sum = 0;
  for (int i = 1; i <= n; ++i)
    if (!vis[i]) q[++top] = Dfs(i, i), sum += (q[top].second);
  pair<int, int> tmp = make_pair(0, 0);
  for (int i = 1; i <= n; ++i)
    if (q[i] > tmp) tmp = q[i];
  if (tmp.first > sum - tmp.second)
    return 0;
  else
    return 1;
}
int main() {
  n = read();
  for (int i = 1; i <= n - 1; ++i)
    e[i].x = read(), e[i].y = read(), e[i].v = read();
  for (int i = 1; i <= n; ++i) xx[i] = read();
  sort(e + 1, e + n, cmp);
  int l = 1, r = 1e9, ans = 0;
  while (l <= r) {
    int mid = l + r >> 1;
    if (check(mid))
      ans = mid, l = mid + 1;
    else
      r = mid - 1;
  }
  writeln(ans);
}