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;
char xB[1 << 15], *xS = xB, *xT = xB;
inline long long read() {
  long long x = 0;
  int f = 1;
  char ch =
      (xS == xT && (xT = (xS = xB) + fread(xB, 1, 1 << 15, stdin), xS == xT)
           ? 0
           : *xS++);
  for (; ch < '0' || ch > '9';
       f = (ch == '-') ? -1 : 1,
       ch = (xS == xT &&
                     (xT = (xS = xB) + fread(xB, 1, 1 << 15, stdin), xS == xT)
                 ? 0
                 : *xS++))
    ;
  for (; ch >= '0' && ch <= '9';
       x = x * 10 + ch - '0',
       ch = (xS == xT &&
                     (xT = (xS = xB) + fread(xB, 1, 1 << 15, stdin), xS == xT)
                 ? 0
                 : *xS++))
    ;
  return f * x;
}
inline void timer() {}
const int N = 3000 + 5;
int n;
int tot;
int head[N];
struct edge {
  int u;
  int v;
  int w;
  bool operator<(const edge &rhs) const { return w < rhs.w; }
};
edge e[N];
int x[N];
int siz[N];
int s[N];
int b[N];
int fa[N];
inline int find(int x) { return x == fa[x] ? x : fa[x] = find(fa[x]); }
inline bool check() {
  for (int i = (0); i < (n); ++i) {
    if (siz[find(i)] > s[find(i)]) {
      return true;
    }
  }
  return false;
}
int main() {
  n = read();
  for (int i = (0); i < (n - 1); ++i) {
    e[i].u = read() - 1;
    e[i].v = read() - 1;
    e[i].w = read();
  }
  sort(e, e + n - 1);
  for (int i = (0); i < (n); ++i) {
    x[i] = read();
  }
  for (int i = (0); i < (n); ++i) {
    fa[i] = i;
    siz[i] = 1;
    for (int j = (0); j < (n); ++j) {
      if (i == j) {
        b[i] = x[i];
        continue;
      }
      s[i] += x[j];
    }
  }
  for (int i = (0), fu, fv; i < (n - 1); ++i) {
    edge p = e[i];
    fu = find(p.u);
    fv = find(p.v);
    fa[fv] = fu;
    siz[fu] += siz[fv];
    s[fu] -= b[fv];
    b[fu] += b[fv];
    if (check()) {
      printf("%d\n", e[i].w);
      return 0;
    }
  }
  printf("%d\n", e[n - 2].w);
  return 0;
}