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 sys import maxint

N = input()
pars = map(int,raw_input().split())
S = map(int,raw_input().split())

T = {1: (None, set())}

for n in range(2,N+1) :
    T[n] = (pars[n-2], set())
    T[pars[n-2]][1].add(n)

def dfs(S, A) :

    st = [1]
    while st :
        cur = st.pop()
        if not T[cur][1] : continue
        if S[cur-1] < 0  : 
            # solve the things
            base = S[T[cur][0]-1] # parent S

            mn = maxint
            for child in T[cur][1] :
                mn = min(mn, S[child-1])
            
            A[cur-1] = mn - base

            for child in T[cur][1] :
                A[child-1] = S[child-1] - mn
        
        for nxt in T[cur][1] : 
            #dfs(nxt, S, A)
            st.append(nxt)

A = [0]*N
A[0] = S[0]
dfs(S, A)

valid = True
ans = 0
for a in A :
    if a < 0 : 
        valid = False
        break
    ans += a

if valid : print ans
else : print -1
#for t in T.items() : print t