Quite recently a creative student Lesha had a lecture on trees. After the lecture Lesha was inspired and came up with the tree of his own which he called a *k*-tree.

A *k*-tree is an infinite rooted tree where:
 -  each vertex has exactly *k* children; -  each edge has some weight; -  if we look at the edges that goes from some vertex to its children (exactly *k* edges), then their weights will equal 1, 2, 3, ..., *k*. 
The picture below shows a part of a 3-tree.

 

 <img class="tex-graphics" src="https://espresso.codeforces.com/61ac5ac17c4d2f220968eff8fb59bb3c8dc77edf.png" style="max-width: 100.0%;max-height: 100.0%;"/>

 

Help Dima find an answer to his question. As the number of ways can be rather large, print it modulo 1000000007 (10<sup class="upper-index">9</sup> + 7).
A single line contains three space-separated integers: *n*, *k* and *d* (1 ≤ *n*, *k* ≤ 100; 1 ≤ *d* ≤ *k*).
def dfs1(u, p):
    f[u] = w[u]
    for v in g[u]:
        if v == p:
            continue
        dfs1(v, u)
        f[u] += f[v]

def dfs2(u, p):
    ans += cnt[f[u] - t]
    cnt[f[u]] += 1
    for v in g[u]:
        if v == p:
            continue
        dfs2(v, u)
    cnt[f[u]] -= 1

n = int(input())
w = [int(x) for x in input().split()]
g = [[] for i in range(n)]
for i in range(n - 1):
    u, v = map(int, input().split())
    u -= 1
    v -= 1
    g[u].append(v)
    g[v].append(u)

t = int(input())

f = [0] * n
dfs1(0, -1)

cnt = [0] * (f[0] + 1)
ans = 0
dfs2(0, -1)

print(ans)