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 = int(input())
p = list(map(int, input().split()))
s = list(map(int, input().split()))

clds, pars = defaultdict(list), [0] * (n + 1)
for i in range(n - 1):
    cur, par = i + 2, p[i]
    clds[p[i]].append(cur)
    pars[cur] = par

ans, ok = s[0], True
i, q = 0, [1]
while i < len(q):
    cur = q[i]
    i += 1
    par_v = s[pars[cur] - 1]
    if cur in clds:
        mmin, summ, k = float('inf'), 0, 0
        for item in clds[cur]:
            mmin = min(mmin, s[item - 1])
            summ += s[item - 1]
            k += 1
            q.append(item)
        if s[cur - 1] != -1:
            continue
        if mmin < par_v:
            ok = False
            break
        ans -= par_v
        ans += (summ - (k - 1) * mmin)
if ok:
    print(ans)
else:
    print(-1)
