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 long long maxn = 1e5 + 500;
const long long mod = 1e9 + 7;
const long long inf = 1e9 + 500;
vector<pair<long long, long long> > ger[maxn];
long long x[maxn];
long long n;
long long par[maxn];
long long summ = 0;
long long sa[maxn];
long long sb[maxn];
long long find_par(long long a) {
  if (par[a] == a) {
    return a;
  }
  par[a] = find_par(par[a]);
  return par[a];
}
void merg(long long a, long long b) {
  a = find_par(a);
  b = find_par(b);
  par[a] = b;
}
bool good(long long m) {
  for (long long i = 0; i < n; i++) {
    par[i] = i;
    sa[i] = 0;
    sb[i] = 0;
  }
  for (long long i = 0; i < n; i++) {
    for (auto e : ger[i]) {
      if (e.second < m) {
        merg(i, e.first);
      }
    }
  }
  for (long long i = 0; i < n; i++) {
    sa[find_par(i)]++;
    sb[find_par(i)] += x[i];
  }
  for (long long i = 0; i < n; i++) {
    if (sa[i] > 0) {
      if (summ - sb[i] < sa[i]) {
        return 0;
      }
    }
  }
  return 1;
}
int main() {
  cin >> n;
  for (long long i = 1; i < n; i++) {
    long long v, u, w;
    cin >> v >> u >> w;
    v--;
    u--;
    ger[u].push_back(make_pair(v, w));
    ger[v].push_back(make_pair(u, w));
  }
  for (long long i = 0; i < n; i++) {
    cin >> x[i];
    summ += x[i];
  }
  long long b = 0;
  long long e = 100000;
  while (e - b > 1) {
    long long mid = (e + b) / 2;
    if (good(mid)) {
      b = mid;
    } else {
      e = mid;
    }
  }
  cout << b;
}