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 double PI = 3.14159265359;
const long long MOD = (long long)998244353ll;
const long long MAXN = (long long)3e3 + 10;
const long long INF = (long long)2242545357980376863;
const long double EPS = (long double)1e-8;
long long n, par[MAXN], w[MAXN], sz[MAXN], wsz[MAXN];
vector<pair<pair<long long, long long>, long long> > E;
long long get_par(long long node) {
  if (par[node] == node) return node;
  par[node] = get_par(par[node]);
  return par[node];
}
void merge(long long u, long long v) {
  u = get_par(u);
  v = get_par(v);
  if (u == v) return;
  par[u] = v;
  sz[v] += sz[u];
  wsz[v] += wsz[u];
  return;
}
vector<pair<long long, long long> > v;
bool valid() {
  long long ind = 0;
  long long wsum = 0;
  long long szsum = 0;
  for (int i = 0; i < v.size(); i++) {
    if (v[i].first > v[ind].first) {
      ind = i;
    }
    szsum += v[i].first;
    wsum += v[i].second;
  }
  if (2 * v[ind].first <= szsum) return true;
  if (wsum - v[ind].second >= v[ind].first) return true;
  return false;
}
bool check(long long weight) {
  fill(sz, sz + MAXN, 1);
  for (int i = 0; i < MAXN; i++) wsz[i] = w[i];
  for (int i = 1; i < MAXN; i++) par[i] = i;
  v.clear();
  for (auto edge : E) {
    if (edge.second < weight) {
      merge(edge.first.first, edge.first.second);
    }
  }
  for (int i = 1; i <= n; i++) {
    if (get_par(i) == i) {
      v.push_back({sz[i], wsz[i]});
    }
  }
  return valid();
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  long long u, v, we;
  for (int i = 1; i < n; i++) {
    cin >> u >> v >> we;
    E.push_back({{u, v}, we});
  }
  for (int i = 1; i <= n; i++) cin >> w[i];
  long long dw = 0;
  long long up = 10001;
  long long mid;
  while (dw + 1 < up) {
    mid = (dw + up) / 2;
    if (check(mid))
      dw = mid;
    else
      up = mid;
  }
  cout << dw << '\n';
  return 0;
}