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().strip())
p = [0 for i in range(n)]
p[0] = -1
index = 1
for i in input().strip().split():
    p[index] = int(i) - 1
    index += 1

min_ = [100000000000 for i in range(n)]
s = []
for i in input().strip().split():
    val = int(i)
    s.append(val)

for i in range(1,n):
    if s[i] >= 0 and s[p[i]] < 0:
        min_[p[i]] = min(min_[p[i]],s[i])

for i in range(1,n):
    if s[i] < 0:
        got = min_[i]
        if got == 100000000000:
            got = s[p[i]]
        s[i] = got
res = s[0]
impossible = False
for i in range(1,n):
    if s[i] < s[p[i]]:
        impossible = True
        break
    res += s[i] - s[p[i]]
if impossible:
    print(-1)
else:
    print(res)
