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 maxn = 100 * 1000 + 100;
const int oo = 1000 * 1000 * 1000;
int n;
vector<pair<int, int> > adj[maxn];
int x[maxn];
bool mark[maxn];
vector<pair<int, int> > sz;
long long cnt1, cnt2;
int bin(int f, int e);
bool chk(int m);
void dfs(int v, int m);
int main() {
  ios ::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    u--, v--;
    adj[u].push_back(pair<int, int>(w, v));
    adj[v].push_back(pair<int, int>(w, u));
  }
  for (int i = 0; i < n; i++) cin >> x[i];
  int ans = bin(0, oo);
  cout << ans << '\n';
  return 0;
}
int bin(int f, int e) {
  if (e - f < 2) return f;
  int mid = (f + e) / 2;
  if (chk(mid)) return bin(mid, e);
  return bin(f, mid);
}
bool chk(int m) {
  fill(mark, mark + n, false);
  sz.resize(0);
  for (int i = 0; i < n; i++)
    if (!mark[i]) {
      cnt1 = 0, cnt2 = 0;
      dfs(i, m);
      sz.push_back(pair<int, int>(cnt1, cnt2));
    }
  long long sum = 0;
  for (int i = 0; i < sz.size(); i++) sum += sz[i].second;
  for (int i = 0; i < sz.size(); i++)
    if (sum - sz[i].second < sz[i].first) return false;
  return true;
}
void dfs(int v, int m) {
  mark[v] = true;
  cnt1++, cnt2 += x[v];
  for (pair<int, int> u : adj[v])
    if (!mark[u.second] && u.first < m) dfs(u.second, m);
}