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 n, u, v, w;
struct node {
  int l, r, v;
  bool operator<(const node& other) const { return v < other.v; }
} edge[101000];
struct no {
  int fa, val, l, r, num, son;
} wjr[201000];
int to[201000], cnt;
inline int find(int x) { return x == to[x] ? x : to[x] = find(to[x]); }
void kruscal() {
  for (int i = 1; i < n; ++i) {
    int t1 = find(edge[i].l), t2 = find(edge[i].r);
    if (t1 != t2) {
      cnt++;
      to[t1] = to[t2] = cnt;
      wjr[cnt].l = t1;
      wjr[cnt].r = t2;
      wjr[t1].fa = wjr[t2].fa = cnt;
      wjr[cnt].val = edge[i].v;
      wjr[cnt].num = wjr[t1].num + wjr[t2].num;
      wjr[cnt].son = wjr[t1].son + wjr[t2].son;
    }
  }
}
int fa[201000][24];
inline void dfs(int x, int last) {
  fa[x][0] = last;
  for (int i = 1; i <= 20; ++i) {
    fa[x][i] = fa[fa[x][i - 1]][i - 1];
  }
  if (wjr[x].l) dfs(wjr[x].l, x);
  if (wjr[x].r) dfs(wjr[x].r, x);
}
bool f[201000];
int maxn;
inline int check(int x) {
  memset(f, 0, sizeof(f));
  for (int i = 1; i <= n; ++i) {
    int now = i;
    for (int j = 20; j >= 0; --j) {
      int nxt = fa[now][j];
      if (nxt && wjr[nxt].val < x) {
        now = fa[now][j];
      }
    }
    if (!f[now] && now > n) {
      f[now] = 1;
      if (wjr[now].son > maxn - wjr[now].num) return 0;
    }
  }
  return 1;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; ++i) {
    scanf("%d%d%d", &u, &v, &w);
    edge[i].l = u;
    edge[i].r = v;
    edge[i].v = w;
  }
  for (int i = 1; i <= n; ++i)
    scanf("%d", &wjr[i].num), wjr[i].son = 1, maxn += wjr[i].num;
  sort(edge + 1, edge + n);
  for (int i = 1; i <= 2 * n + 5; ++i) to[i] = i;
  cnt = n;
  kruscal();
  dfs(cnt, 0);
  int l = 0, r = n, ans = 0;
  edge[0].v = 0;
  edge[n].v = edge[n - 1].v + 1000;
  while (l < r) {
    int mid = (l + r) >> 1;
    if (check(edge[mid].v)) {
      l = mid + 1;
      ans = edge[mid].v;
    } else {
      r = mid;
    }
  }
  printf("%d\n", ans);
  return 0;
}