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 = 5e5 + 10;
int n;
int ans;
int u[maxn];
int v[maxn];
int w[maxn];
int O[maxn];
int par[maxn];
int num[maxn];
void in();
void solve();
int dsu(int, int);
bool cmp(int, int);
int root(int);
void out();
int main() {
  ios::sync_with_stdio(false);
  cout.tie(0);
  cin.tie(0);
  in();
  solve();
  out();
}
void in() {
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    cin >> u[i] >> v[i] >> w[i];
    u[i]--;
    v[i]--;
  }
  for (int i = 0; i < n; i++) cin >> num[i];
}
void solve() {
  int sum = 0;
  for (int i = 0; i < n; i++) {
    sum += num[i];
    par[i] = -1;
    O[i] = i;
  }
  sort(O, O + n - 1, cmp);
  ans = w[O[0]];
  int ded = 0;
  for (int i = 0; i < n - 1; i++) {
    int x = dsu(u[O[i]], v[O[i]]);
    if (par[x] + sum - num[x] < 0) ded = 1;
    if (!ded) ans = w[O[i + 1]];
  }
}
int dsu(int u, int v) {
  u = root(u);
  v = root(v);
  if (par[u] < par[v]) swap(u, v);
  par[u] += par[v];
  num[u] += num[v];
  par[v] = u;
  return u;
}
int root(int u) { return par[u] < 0 ? u : par[u] = root(par[u]); }
bool cmp(int i, int j) { return w[i] < w[j]; }
void out() { cout << ans; }