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())
parents = list(map(int, input().split(' ')))
S = list(map(int, input().split(' ')))

children = [[] for i in range(n)]
for i, parent in enumerate(parents):
    children[parent-1].append(i+2)

A = S[:]
for i in range(1, len(A)):
    if A[i] != -1:
        A[i] -= S[parents[parents[i-1]-2]-1]
        if A[i] < 0:
            print(-1)
            exit(0)

for i in range(1, len(A)):
    if A[i] == -1:
        if len(children[i]) == 0:
            A[i] = 0
        else:
            A[i] = min([A[j-1] for j in children[i]])
        for j in children[i]:
            A[j-1] -= A[i]

print(sum(A))