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 p_1, p_2, ..., p_{n} (1 ≤ p_{i} ≤ n), DZY defines f(p) as $\operatorname{min}_{i = 1}^{n} g(i, p_{i})$. 
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 x_{j} 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 a_{i}, b_{i}, c_{i} (1 ≤ a_{i}, b_{i} ≤ n; 1 ≤ c_{i} ≤ 10000), denoting an edge between a_{i} and b_{i} with length c_{i}. 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 x_{j} (1 ≤ x_{j} ≤ 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].
n = int(input())
edges = [[int(x) for x in input().split()] for i in range(n-1)]
edges = sorted(edges)
use_count = [0]+[int(input()) for i in range(n)]
lo,hi = 0,10000
def getpar(par,u):
    if par[par[u]] == par[u]:
        return par[u]
    par[u] = getpar(par,par[u])
    return par[u]
def unite(par,sz,use,u,v):
    u = getpar(par,u)
    v = getpar(par,v)
    par[u] = v
    sz[v] += sz[u]
    use[v] += use[u]
def solve(fp):
    par = [i for i in range(n+1)]
    sz = [1 for i in range(n+1)]
    use = [use_count[i] for i in range(n+1)]
    for edge in edges:
        if edge[2] < fp:
            unite(par,sz,use,edge[0],edge[1])
    total_use = sum(use_count)
    for i in range(n+1):
        p = getpar(par,i)
        if(p == i):
            if(total_use - use[p] < sz[p]):
                return False
    return True
while lo < hi:
    mid = (lo+hi+1)//2
    if solve(mid):
        lo = mid
    else:
        hi = mid-1
print(lo)

n = int(input())
edges = [[int(x) for x in input().split()] for i in range(n-1)]
edges = sorted(edges)
use_count = [0]+[int(input()) for i in range(n)]
lo,hi = 0,10000
def getpar(par,u):
    if par[par[u]] == par[u]:
        return par[u]
    par[u] = getpar(par,par[u])
    return par[u]
def unite(par,sz,use,u,v):
    u = getpar(par,u)
    v = getpar(par,v)
    par[u] = v
    sz[v] += sz[u]
    use[v] += use[u]
def solve(fp):
    par = [i for i in range(n+1)]
    sz = [1 for i in range(n+1)]
    use = [use_count[i] for i in range(n+1)]
    for edge in edges:
        if edge[2] < fp:
            unite(par,sz,use,edge[0],edge[1])
    total_use = sum(use_count)
    for i in range(n+1):
        p = getpar(par,i)
        if(p == i):
            if(total_use - use[p] < sz[p]):
                return False
    return True
while lo < hi:
    mid = (lo+hi+1)//2
    if solve(mid):
        lo = mid
    else:
        hi = mid-1
print(lo)
