Mitya has a rooted tree with n vertices indexed from 1 to n, where the root has index 1. Each vertex v initially had an integer number a_v ≥ 0 written on it. For every vertex v Mitya has computed s_v: the sum of all values written on the vertices on the path from vertex v to the root, as well as h_v — the depth of vertex v, which denotes the number of vertices on the path from vertex v to the root. Clearly, s_1=a_1 and h_1=1.

Then Mitya erased all numbers a_v, and by accident he also erased all values s_v for vertices with even depth (vertices with even h_v). Your task is to restore the values a_v for every vertex, or determine that Mitya made a mistake. In case there are multiple ways to restore the values, you're required to find one which minimizes the total sum of values a_v for all vertices in the tree.

Input

The first line contains one integer n — the number of vertices in the tree (2 ≤ n ≤ 10^5). The following line contains integers p_2, p_3, ... p_n, where p_i stands for the parent of vertex with index i in the tree (1 ≤ p_i < i). The last line contains integer values s_1, s_2, ..., s_n (-1 ≤ s_v ≤ 10^9), where erased values are replaced by -1.

Output

Output one integer — the minimum total sum of all values a_v in the original tree, or -1 if such tree does not exist.

Examples

Input


5
1 1 1 1
1 -1 -1 -1 -1


Output


1


Input


5
1 2 3 1
1 -1 2 -1 -1


Output


2


Input


3
1 2
2 -1 1


Output


-1
n=int(input())
p=list(map(int,input().split()))
s=list(map(int,input().split()))
g=[[] for i in range(n)]
for i in range(n-1):
  g[p[i]-1].append(i+1)
  g[i+1].append(p[i]-1)
c=[0]*n
q=[(-1,0,0)]
c[0]=s[0]
p=[0]+p
while q:
  par,ver,h=q.pop()
  if h%2==1:
    if len(g[ver])==1:
      c[ver]=0
      continue
    m=10**18
    for to in g[ver]:
      if par==to:
        continue
      m=min(m,s[to]-s[par])
      q.append((ver,to,h+1))
    if m<0:
      print(-1)
      exit()
    c[ver]=m
  else:
    if h>=2:
      c[ver]=s[ver]-s[p[par]-1]-c[par]
      if c[ver]<0:
        print(-1)
        exit()
    for to in g[ver]:
      if par!=to:
        q.append((ver,to,h+1))
print(sum(c))