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 int read() {
  int 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(int x) {
  if (x < 0) putchar('-'), x = -x;
  if (x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
inline void writeln(int x) {
  write(x);
  puts("");
}
const int N = 1e5 + 5;
struct edge {
  int u, v, w;
} e[N << 1];
int n, m, c[N], Sum;
inline void init() {
  n = read();
  for (int i = 1; i < n; i++) {
    e[i] = (edge){read(), read(), read()};
  }
  for (int i = 1; i <= n; i++) {
    c[i] = read();
    Sum += c[i];
  }
}
inline bool cmp(edge A, edge B) { return A.w < B.w; }
int tmp, fa[N], sz[N], sum[N];
int getfa(int x) { return (fa[x] == x) ? x : fa[x] = getfa(fa[x]); }
inline void Union(int u, int v) {
  int p = getfa(u), q = getfa(v);
  if (p != q) {
    fa[p] = q;
    sz[q] += sz[p];
  }
}
inline bool judge(int mid) {
  for (int i = 1; i <= n; i++) fa[i] = i, sz[i] = c[i];
  memset(sum, 0, sizeof sum);
  tmp = 0;
  for (int i = 1; i <= m; i++) {
    if (e[i].w < mid) {
      tmp = i;
      Union(e[i].u, e[i].v);
    } else {
      break;
    }
  }
  for (int i = 1; i <= n; i++) {
    fa[i] = getfa(i);
    sum[fa[i]]++;
  }
  for (int i = 1; i <= n; i++) {
    if (sum[i]) {
      int other = Sum - sz[i];
      if (other < sum[i]) {
        return 0;
      }
    }
  }
  return 1;
}
inline void solve() {
  if (n == 1) {
    puts("0");
    return;
  }
  m = n - 1;
  sort(e + 1, e + m + 1, cmp);
  int l = 1, r = 1e9;
  while (l < r) {
    int mid = (l + r + 1) >> 1;
    if (judge(mid))
      l = mid;
    else
      r = mid - 1;
  }
  writeln(l);
}
int main() {
  init();
  solve();
  return 0;
}