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 mod = 1000 * 1000 * 1000 + 7;
const int INF = 1000 * 1000 * 1000;
const long long LINF = (long long)INF * INF;
int n, efrom[3030], eto[3030], ew[3030];
int ed[3030];
int bnd[3030];
int dpar[3030];
int cap[3030], sz[3030];
int getroot(int x) { return (dpar[x] == x ? x : dpar[x] = getroot(dpar[x])); }
void merge(int x, int y) {
  x = getroot(x);
  y = getroot(y);
  if (x == y) return;
  sz[y] += sz[x];
  cap[y] += cap[x];
  dpar[x] = y;
}
bool check(int mid) {
  iota(dpar, dpar + 3030, 0);
  fill(sz, sz + 3030, 1);
  copy(bnd, bnd + 3030, cap);
  for (int i = 0; i < n - 1 && ew[ed[i]] < mid; i++) {
    merge(efrom[ed[i]], eto[ed[i]]);
  }
  int capsum = 0;
  for (int i = (1); i < (n + 1); i++)
    if (dpar[i] == i) capsum += cap[i];
  for (int i = (1); i < (n + 1); i++)
    if (dpar[i] == i && sz[i] > capsum - cap[i]) return false;
  return (capsum >= n);
}
int32_t main(void) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n;
  for (int i = (0); i < (n - 1); i++) {
    int x, y, w;
    cin >> x >> y >> w;
    efrom[i] = x;
    eto[i] = y;
    ew[i] = w;
    ed[i] = i;
  }
  for (int i = (1); i < (n + 1); i++) cin >> bnd[i];
  sort(ed, ed + n - 1, [&](int x, int y) { return ew[x] < ew[y]; });
  int l = 0, r = 1e5;
  while (r - l > 1) {
    int mid = (r + l) / 2;
    if (check(mid))
      l = mid;
    else
      r = mid;
  }
  cout << l << '\n';
  return 0;
}