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 N = 3010;
struct node {
  int x, y, z;
} e[N];
bool cmp(node x, node y) { return x.z < y.z; }
int x[N], fa[N], siz[N], siz2[N];
int fid(int x) {
  if (x == fa[x]) return x;
  return fa[x] = fid(fa[x]);
}
int read() {
  int x = 0, f = 1;
  char c = 0;
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - 48;
  return x * f;
}
int main() {
  int n = read(), sum = 0;
  for (int i = 1; i < n; i++) e[i] = (node){read(), read(), read()};
  for (int i = 1; i <= n; i++) {
    x[i] = read();
    fa[i] = i, siz[i] = 1;
    siz2[i] = x[i];
    sum += x[i];
  }
  sort(e + 1, e + n, cmp);
  for (int i = 1; i < n; i++) {
    int x = e[i].x, y = e[i].y;
    int fx = fid(x), fy = fid(y);
    if (fx == fy) continue;
    if (siz[fx] > siz[fy]) swap(fx, fy);
    fa[fx] = fy;
    siz[fy] += siz[fx];
    siz2[fy] += siz2[fx];
    if (sum - siz2[fy] < siz[fy]) {
      cout << e[i].z << "\n";
      return 0;
    }
  }
  cout << e[n - 1].z << "\n";
  return 0;
}