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>
#pragma comment(linker, "/STACK:256000000")
using namespace std;
const int N = 3100;
struct edge {
  int a, b, c;
};
bool operator<(const edge& l, const edge& r) {
  if (l.c != r.c) return l.c < r.c;
  if (l.a != r.a) return l.a < r.a;
  return l.b < r.b;
}
int n, cap[N];
edge e[N];
int prnt[N], rnk[N], nd[N], fr[N];
int find(int x) { return x == prnt[x] ? x : prnt[x] = find(prnt[x]); }
void unite(int x, int y) {
  x = find(x);
  y = find(y);
  if (x == y) return;
  if (rnk[x] >= rnk[y]) {
    prnt[y] = x;
    nd[x] += nd[y];
    fr[x] += fr[y];
  } else {
    prnt[x] = y;
    nd[y] += nd[x];
    fr[y] += fr[x];
  }
  if (rnk[x] == rnk[y]) ++rnk[x];
}
void init() {
  for (int i = 0; i <= n; ++i) {
    prnt[i] = i;
    rnk[i] = 0;
    nd[i] = 1;
    fr[i] = cap[i];
  }
}
bool check(int f) {
  init();
  int i = 0;
  for (; i < n - 1 && e[i].c < f; ++i) unite(e[i].a, e[i].b);
  int tl = 0;
  for (int i = 1; i <= n; ++i)
    if (prnt[i] == i) tl += fr[i];
  for (int i = 1; i <= n; ++i)
    if (prnt[i] == i) {
      if (nd[i] > tl - fr[i]) return false;
    }
  return true;
}
int main() {
  scanf("%d", &n);
  int l = 0, r = 0;
  for (int i = 0; i < n - 1; ++i) {
    scanf("%d %d %d", &e[i].a, &e[i].b, &e[i].c);
    r = max(r, e[i].c + 1);
  }
  for (int i = 1; i <= n; ++i) scanf("%d", cap + i);
  sort(e, e + n - 1);
  while (r - l > 1) {
    int mid = (l + r) / 2;
    if (check(mid))
      l = mid;
    else
      r = mid;
  }
  cout << l << endl;
  return 0;
}