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;
int fa[3010], cnt[3010], x[3010], hop[3010];
int to[2 * 3010], nxt[2 * 3010], wei[2 * 3010], head[3010], edge;
int val[3010], N;
void add_edge(int u, int v, int w) {
  to[edge] = v;
  nxt[edge] = head[u];
  wei[edge] = w;
  head[u] = edge++;
  to[edge] = u;
  nxt[edge] = head[v];
  wei[edge] = w;
  head[v] = edge++;
  return;
}
int find_fa(int u) {
  if (u == fa[u]) return u;
  return fa[u] = find_fa(fa[u]);
}
void dfs(int u, int bound, int f) {
  for (int ID = head[u]; ID != -1; ID = nxt[ID]) {
    int v = to[ID];
    if (v == f) continue;
    if (wei[ID] < bound) fa[find_fa(u)] = find_fa(v);
    dfs(v, bound, u);
  }
  return;
}
bool calc(int t) {
  for (int i = 0; i < N; i++) fa[i] = i;
  for (int i = 0; i < N; i++) cnt[i] = 0;
  for (int i = 0; i < N; i++) hop[i] = 0;
  dfs(0, t, -1);
  for (int i = 0; i < N; i++) fa[i] = find_fa(i);
  for (int i = 0; i < N; i++) cnt[fa[i]]++;
  for (int i = 0; i < N; i++) hop[fa[i]] += x[i];
  int mxcnt = fa[0];
  int sum = 0;
  for (int i = 0; i < N; i++)
    if (fa[i] == i) {
      sum += hop[i];
      if (cnt[i] > cnt[mxcnt]) mxcnt = i;
    }
  if (sum - hop[mxcnt] >= cnt[mxcnt]) return true;
  return false;
}
int main() {
  scanf("%d", &N);
  memset(head, 0xff, sizeof(head));
  int ct = 0;
  int u, v, w, fir, se, mid;
  for (int i = 0; i < N - 1; i++) {
    scanf("%d%d%d", &u, &v, &w);
    u--;
    v--;
    add_edge(u, v, w);
    val[ct++] = w;
  }
  for (int i = 0; i < N; i++) scanf("%d", x + i);
  sort(val, val + ct);
  ct = unique(val, val + ct) - val;
  fir = 0;
  se = ct;
  while (se - fir > 1) {
    mid = (se + fir) >> 1;
    if (calc(val[mid]))
      fir = mid;
    else
      se = mid;
  }
  printf("%d\n", val[fir]);
  return 0;
}