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 = 200100;
long long t[MAXN], g[MAXN], pai[MAXN];
struct str {
  long long l, r, w;
};
str are[MAXN];
bool comp(str a, str b) { return a.w < b.w; }
bool comp2(long long x, long long y) { return t[x] > t[y]; }
long long find(long long x) {
  if (pai[x] == x) return pai[x];
  return pai[x] = find(pai[x]);
}
void join(long long a, long long b) {
  long long x = find(a), y = find(b);
  g[y] += g[x];
  t[y] += t[x];
  pai[x] = y;
}
bool vis[MAXN];
signed main() {
  memset(g, 0, sizeof g);
  long long n;
  cin >> n;
  for (long long i = 1; i <= n; i++) t[i] = 1, pai[i] = i;
  for (long long i = 1; i < n; i++) {
    long long x, y, z;
    cin >> x >> y >> z;
    are[i].l = x, are[i].r = y, are[i].w = z;
  }
  long long soma = 0;
  for (long long i = 1; i <= n; i++) {
    long long x;
    cin >> x;
    g[i] = x - 1;
    soma += x - 1;
  }
  sort(are + 1, are + n, comp);
  long long resp = 0;
  for (long long i = 0; i < n - 1; i++) {
    if (i != 0) join(are[i].l, are[i].r);
    memset(vis, false, sizeof vis);
    vector<long long> vec;
    for (long long j = 1; j <= n; j++) {
      long long x = find(j);
      if (vis[x] == true) continue;
      vis[x] = true;
      vec.push_back(x);
    }
    sort(vec.begin(), vec.end(), comp2);
    if (t[vec[0]] <= n - t[vec[0]] + soma - g[vec[0]])
      resp = are[i + 1].w;
    else
      break;
  }
  cout << resp << endl;
}