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
from collections import defaultdict
n=input()
p=[0,0]+map(int,raw_input().split())
s=[0]+map(int,raw_input().split())
d=defaultdict(list)
for i in range(2,n+1):
    d[p[i]].append(i)
q=[1]
ans=0
while q:
    x=q.pop()
    #print q
    mn=1000000000000000
    for i in d[x]:
        mn=min(mn, s[i])
        q.append(i)
        
    if s[x]==-1:
        s[x]=mn
        #print x,mn
        if mn<s[p[x]] and mn!=1000000000000000:
            print -1
            exit()
        elif mn==1000000000000000:
            mn=s[p[x]]
        ans+=(mn-s[p[x]])
    else:
        if s[x]<s[p[x]]:
            print -1
            exit()
        else:
            #print x,s[x]-s[p[x]]
            ans+=(s[x]-s[p[x]])
print ans
