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 int MOD = (int)1e9 + 7;
const int MAXN = (int)3007;
const int infint = (int)1e9;
const long long inf = (long long)1e18;
const int bound = 10007;
int n, x[MAXN], par[MAXN], sz[MAXN], bip[MAXN];
struct edge {
  int u, v, w;
  edge(int a, int b, int c) { u = a, v = b, w = c; }
};
vector<edge> e;
bool cmp(edge E1, edge E2) { return E1.w < E2.w; }
int get(int v) { return par[v] < 0 ? v : par[v] = get(par[v]); }
void merge(int u, int v) {
  if ((u = get(u)) == (v = get(v))) return;
  if (par[u] > par[v]) swap(u, v);
  par[u] += par[v];
  par[v] = u;
}
bool hall() {
  int sm = 0;
  for (int i = 1; i <= n; i++) sm += bip[i];
  if (sm < n) return 0;
  for (int i = 1; i <= n; i++)
    if (sz[i] > sm - bip[i]) return 0;
  return 1;
}
bool check(int mid) {
  memset(par, -1, sizeof par);
  for (auto E : e)
    if (E.w < mid)
      merge(E.u, E.v);
    else
      break;
  memset(sz, 0, sizeof sz);
  memset(bip, 0, sizeof bip);
  for (int i = 1; i <= n; i++) sz[get(i)]++, bip[get(i)] += x[i];
  return hall();
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    edge nw(u, v, w);
    e.push_back(nw);
  }
  sort(e.begin(), e.end(), cmp);
  for (int i = 1; i <= n; i++) cin >> x[i];
  int L = 0, R = bound;
  while (R - L > 1) {
    int mid = (L + R) >> 1;
    if (check(mid))
      L = mid;
    else
      R = mid;
  }
  cout << L;
}