T1
n, a, b, c = map(int, input() .split())
cache = dict()
def dfs(i, a ,b, g):
if (i,a, b,c) in cache:
return 0
ans = int(a + b > c and a + c > b and b + c > a)
if n < i:
cache[(i,a,b,c)] = ans
return ans
if c > i:
ans += dfs(i + 1,a, b, c - i)
if b > i:
ans += dfs(i + 1,a, b - i, c)
if a > i:
ans += dfs(i + 1,a - i, b, c)
cache[(i, a,b,c)] = ans
return ans
print(dfs(1,a, b, c))
T2 换根dp，先dfs一次算出根为0时候的答案，记录一下每个节点的子树的大小，然后换根，计算的时候可以O(1)转移根节点
def dfs1(root):
sz = 1
for to in adj[root]:
if vis[to]:
continue
vis[to] = 1
sz2, ans2 = dfs1(to)
sz += sz2
ans[root] += ans2
ans[root] += sz - 1
szv[root] = sz
return sz, ans[root]
def dfs2(root):
for to in adj[root]:
if vis[to]:
continue
vis[to] = 1
t = ans[to]
ans[to] += n - szv[to]
ans[to] += ans[root] - t - szv[to]
dfs2(to)
n, m = map(int, input().split())
adj = [[] for _ in range(n)]
p = [-1] * n
szv = [0] * n
ans = [0] * n
xv = list(map(int,input().split()))
for i in range(1, n):
x = xv[i-1]-1
p[i] = x
adj[i].append(x)
adj[x].append(i)
vis = [0] * n
vis[0] = 1
dfs1(0)
vis = [0] * n
vis[0] = 1
dfs2(0)
xv = list(map(int,input().split()))
for i in range(m):
v = xv[i] - 1
print(ans[v], end=" ")
#滴滴##笔试#
