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;
long long int MOD = 1e9 + 7;
const int N = 3005, M = 1e5 + 10, NM1 = N - 1;
long long int OO = 1e18 + 8;
int oo = 1e9 + 8;
void readFile() {}
struct DSU {
  vector<int> par, sz;
  DSU(int n) {
    par.resize(n), sz.resize(n), iota(par.begin(), par.end(), 0),
        fill(sz.begin(), sz.end(), 1);
  }
  int leader(int u) { return (par[u] == u) ? u : par[u] = leader(par[u]); }
  bool join(int u, int v) {
    int p1 = leader(u), p2 = leader(v);
    if (p1 == p2) return false;
    par[p2] = p1;
    sz[p1] += sz[p2];
    return true;
  }
} dsu(1);
int n, m, u, v, x[N], edgeFrom[N], edgeTo[N], edgeCost[N], sorted[N];
int main() {
  readFile();
  scanf("%d", &n);
  dsu = DSU(n + 1);
  if (n == 1) return puts("0"), 0;
  for (int i = 0; i < n - 1; i++) {
    scanf("%d%d%d", edgeFrom + i, edgeTo + i, edgeCost + i);
    sorted[i] = i;
  }
  sort(sorted, sorted + n - 1,
       [](int a, int b) { return edgeCost[a] < edgeCost[b]; });
  int sum = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", x + i);
    sum += x[i];
  }
  for (int i = 0; i < n - 1; i++) {
    int idx = sorted[i];
    u = dsu.leader(edgeFrom[idx]), v = dsu.leader(edgeTo[idx]);
    dsu.join(u, v);
    x[u] += x[v];
    if (dsu.sz[dsu.leader(u)] > sum - x[u])
      return printf("%d\n", edgeCost[idx]), 0;
  }
  printf("%d\n", edgeCost[sorted[n - 2]]);
  return 0;
}