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()))
for i in range(len(p)):
    p[i] = p[i] -1
p.insert(0,-1)
#print(lista)
c = pow(10,9) + 5
mina = [c for i in range(n)]
for i in range(1,n):
    if(s[i] >= 0 and s[p[i]] < 0):
        mina[p[i]] = min(mina[p[i]],s[i])
for i in range(n):
    if(s[i] < 0):
        if(mina[i] == c):
            s[i] = s[p[i]]
        else:
            s[i] = mina[i]
ans = s[0]
flag = 0
for i in range(1,n):
    if(s[i] < s[p[i]]):
        flag = 1
        print(-1)
        break
    ans += s[i] - s[p[i]]

if(flag == 0):
    print(ans)








