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, w = 0;
  char c = getchar();
  while (!isdigit(c)) w |= c == '-', c = getchar();
  while (isdigit(c)) x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
  return w ? -x : x;
}
namespace star {
const int maxn = 1e5 + 10;
int n, m;
int ecnt, head[maxn], nxt[maxn << 1], to[maxn << 1], v[maxn], mid;
inline void addedge(int a, int b, int c) {
  to[++ecnt] = b, nxt[ecnt] = head[a], head[a] = ecnt;
  v[ecnt] = c;
  to[++ecnt] = a, nxt[ecnt] = head[b], head[b] = ecnt;
  v[ecnt] = c;
}
int siz[maxn], fsiz[maxn];
long long sum[maxn], fsum[maxn];
void dfs(int x, int f) {
  siz[x] = 1;
  for (int i = head[x]; i; i = nxt[i]) {
    int u = to[i];
    if (u == f) continue;
    dfs(u, x);
    siz[x] += siz[u];
    sum[x] += sum[u];
  }
}
bool ok;
void dfs1(int x, int top, int fa) {
  for (int i = head[x]; i; i = nxt[i]) {
    int u = to[i];
    if (u == fa) continue;
    if (v[i] >= mid)
      dfs1(u, u, x), fsiz[top] -= siz[u], fsum[top] += sum[u];
    else
      dfs1(u, top, x);
  }
  if (x == top) {
    if ((sum[1] - sum[x]) + fsum[x] < 1LL * fsiz[x]) ok = 0;
  }
}
inline bool check() {
  ok = 1;
  memcpy(fsiz, siz, sizeof fsiz);
  memset(fsum, 0, sizeof fsum);
  dfs1(1, 1, 1);
  return ok;
}
inline void work() {
  n = read();
  int l = 0, r = 0, ans = 0;
  for (int a, b, c, i = 1; i < n; i++)
    a = read(), b = read(), c = read(), r = max(r, c), addedge(a, b, c);
  for (int i = 1; i <= n; i++) sum[i] = read();
  dfs(1, 1);
  while (l <= r) {
    mid = l + r >> 1;
    if (check())
      ans = mid, l = mid + 1;
    else
      r = mid - 1;
  }
  printf("%d", ans);
}
}  // namespace star
signed main() {
  star::work();
  return 0;
}