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;
inline long long mod(long long n, long long m) {
  long long ret = n % m;
  if (ret < 0) ret += m;
  return ret;
}
long long gcd(long long a, long long b) {
  return (b == 0LL ? a : gcd(b, a % b));
}
long long exp(long long a, long long b, long long m) {
  if (b == 0LL) return 1LL;
  if (b == 1LL) return mod(a, m);
  long long k = mod(exp(a, b / 2, m), m);
  if (b & 1LL) {
    return mod(a * mod(k * k, m), m);
  } else
    return mod(k * k, m);
}
struct dsu {
  vector<long long> p, ps;
  dsu() {}
  dsu(long long n) {
    p = vector<long long>(n + 1), ps = vector<long long>(n + 1, 1);
    for (long long i = 0; i < (n + 1); ++i) p[i] = i;
  }
  long long find(long long x) { return p[x] == x ? x : p[x] = find(p[x]); }
  bool join(long long x, long long y) {
    x = find(x), y = find(y);
    if (x == y) return 0;
    if (ps[x] > ps[y]) swap(x, y);
    p[x] = y, ps[y] += ps[x];
    return 1;
  }
};
const long long N = 3030;
long long a[N], b[N], c[N];
long long x[N];
long long n;
long long pai[N];
long long sumx[N];
long long sz[N];
long long f(long long x) { return pai[x] = (pai[x] == x ? x : f(pai[x])); }
void join(long long a, long long b) {
  a = f(a), b = f(b);
  if (f(a) == f(b)) return;
  pai[b] = a;
  sumx[a] += sumx[b];
  sz[a] += sz[b];
  return;
}
bool ok(long long mid) {
  long long tot = 0;
  for (long long i = 1; i <= n; i++) {
    pai[i] = i;
    sz[i] = 1;
    sumx[i] = x[i];
    tot += x[i];
  }
  for (long long i = 1; i < n; i++) {
    if (c[i] < mid) {
      join(a[i], b[i]);
    }
  }
  long long flow = 1e9;
  for (long long i = 1; i <= n; i++) {
    if (f(i) != i) continue;
    flow = min(flow, tot + n - sz[f(i)] - sumx[f(i)]);
  }
  return flow >= n;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n;
  for (long long i = 1; i < n; i++) {
    cin >> a[i] >> b[i] >> c[i];
  }
  for (long long i = 1; i <= n; i++) cin >> x[i];
  long long l = 1, r = 100000;
  long long ans = 0;
  while (l <= r) {
    long long mid = (l + r) / 2;
    if (ok(mid)) {
      ans = mid;
      l = mid + 1;
    } else
      r = mid - 1;
  }
  cout << ans << "\n";
}