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 k = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    k = k * 10 + ch - '0';
    ch = getchar();
  }
  return k * f;
}
inline void write(long long x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
inline void writeln(long long x) {
  write(x);
  puts("");
}
const long long N = 1e5 + 10;
long long n, sum = 0;
struct ppap {
  long long x, y, v;
} a[N];
inline bool cmp(ppap a, ppap b) { return a.v < b.v; }
long long fa[N], s[N], b[N];
inline long long find(long long x) {
  if (fa[x] == x) return x;
  return fa[x] = find(fa[x]);
}
signed main() {
  n = read();
  for (long long i = 1; i < n; i++) {
    a[i].x = read();
    a[i].y = read();
    a[i].v = read();
  }
  for (long long i = 1; i <= n; i++) {
    b[i] = read();
    fa[i] = i;
    s[i] = 1;
    sum += b[i];
  }
  sort(a + 1, a + n, cmp);
  long long i = 1;
  for (; i <= n; i++) {
    long long fx = find(a[i].x), fy = find(a[i].y);
    if (fx != fy) {
      fa[fx] = fy;
      s[fy] += s[fx];
      b[fy] += b[fx];
      if (s[fy] > sum - b[fy]) break;
    }
  }
  writeln(a[i].v);
  return 0;
}