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;
const long long MOD = 1e9 + 7;
const long long MAXN = 2e4 + 30;
const long long MINN = 1e3 + 20;
const long long MOD2 = 998244353ll;
const long long INF = 74592896151251;
const long double EPS = 1e-9;
long long GCD(long long a, long long b) { return (b ? GCD(b, a % b) : a); }
long long POW(long long a, long long b) {
  return (!b ? 1 : POW(a, b / 2) * POW(a, b / 2) * (b % 2 ? a : 1));
}
long long X[MAXN], Par[MAXN], SZ[MAXN], n, Sum;
struct Edge {
  long long u, v, w;
} E[MAXN];
inline void input() {
  cin >> n;
  for (int i = 0; i < n - 1; i++) cin >> E[i].u >> E[i].v >> E[i].w;
  for (int i = 1; i <= n; i++) cin >> X[i], Sum += X[i];
}
inline long long Find(long long v) {
  if (v == Par[v]) return v;
  return (Par[v] = Find(Par[v]));
}
inline void Union(long long u, long long v) {
  v = Find(v);
  u = Find(u);
  if (u == v) return;
  Par[v] = u;
  SZ[u] += SZ[v];
  X[u] += X[v];
}
bool CMP(Edge x, Edge y) { return (x.w < y.w); }
inline void Pre_Process() {
  for (int i = 1; i <= n; i++) Par[i] = i, SZ[i] = 1;
}
inline int Main() {
  input();
  Pre_Process();
  sort(E, E + n - 1, CMP);
  for (int i = 0; i < n - 1; i++) {
    E[i].v = Find(E[i].v);
    E[i].u = Find(E[i].u);
    Union(E[i].u, E[i].v);
    if (SZ[E[i].u] > Sum - X[E[i].u]) return cout << E[i].w, 0;
  }
  cout << 0;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  int q;
  q = 1;
  while (q--) Main();
  return 0;
}