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 int MOD = 1000000007;
const long long BIG = 1446803456761533460;
const int Big = 336860180;
stringstream sss;
const long long int maxn = 3010;
long long int n;
long long int cnt[maxn], sz[maxn];
long long int rt[maxn];
pair<long long int, pair<long long int, long long int> > es[maxn];
long long int root(long long int x) {
  return x == rt[x] ? x : rt[x] = root(rt[x]);
}
void MAIN() {
  cin >> n;
  for (long long int i = (0); i < (n - 1); ++i) {
    long long int a, b, c;
    cin >> a >> b >> c;
    --a, --b;
    es[i] = {c, {a, b}};
  }
  long long int sum = 0;
  for (long long int i = (0); i < (n); ++i) cin >> cnt[i], sum += cnt[i];
  sort(es, es + n - 1);
  iota(rt, rt + n, 0);
  fill(sz, sz + n, 1);
  long long int ans = 0;
  for (long long int i = (0); i < (n - 1); ++i) {
    ans = es[i].first;
    long long int ra = root(es[i].second.first), rb = root(es[i].second.second);
    sz[ra] += sz[rb];
    cnt[ra] += cnt[rb];
    rt[rb] = ra;
    if (sum - cnt[ra] < sz[ra]) break;
  }
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cout << fixed << setprecision(10);
  sss << R"(
4
1 2 1
2 3 2
3 4 3
1
1
1
1
    )";
  MAIN();
  return 0;
}