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 N = 100010;
using namespace std;
int n, head[N], k = 0, w[N], kp[N];
long long size[N], sum = 0;
struct M {
  int to, next, val;
} edge[N * 2];
void build(int x, int y, int z) {
  edge[++k].next = head[x];
  edge[k].to = y;
  edge[k].val = z;
  head[x] = k;
}
void find(int x, int f) {
  size[x] = w[x];
  kp[x] = 1;
  for (int i = head[x]; i; i = edge[i].next) {
    int v = edge[i].to;
    if (v == f) continue;
    find(v, x);
    size[x] += size[v];
    kp[x] += kp[v];
  }
}
void dfs(int x, int f, int lp) {
  for (int i = head[x]; i; i = edge[i].next) {
    int v = edge[i].to;
    if (v == f) continue;
    if (edge[i].val >= lp) {
      if (size[v] <= kp[1] - kp[v]) {
        sum += size[v] + kp[v];
        continue;
      } else
        sum += size[1] - size[v] + (kp[1] - kp[v]), dfs(v, x, lp);
    } else
      dfs(v, x, lp);
  }
}
bool check(int lp) {
  sum = 0;
  dfs(1, 1, lp);
  return sum >= n;
}
int main() {
  int x, y, z, l = 1e9, r = 0, ans = 0;
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    scanf("%d%d%d", &x, &y, &z);
    build(x, y, z), build(y, x, z);
    l = min(l, z), r = max(r, z);
  }
  for (int i = 1; i <= n; i++) scanf("%d", &w[i]);
  find(1, 1);
  while (l <= r) {
    int mid = (l + r) / 2;
    if (check(mid))
      ans = mid, l = mid + 1;
    else
      r = mid - 1;
  }
  printf("%d\n", ans);
  return 0;
}