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())

arr = list(map(int,input().strip().split()))

sr = list(map(int,input().strip().split()))

gr = [[] for i in range(n)]

for i in range(len(arr)):
    gr[arr[i]-1].append(i+1)

q = [0]

liv = {0:1}
par = {0:0}
ans = [-1]*(n)

if sr[0]!=-1:
    ans[0]=sr[0]
#print(gr)
while q:
    ne = []

    for u in q:
        #print(q)
        if sr[u]==-1:
            a = []
            for v in gr[u]:
                a.append(sr[v]-sr[par[u]])
            if len(a)==0:
                continue
            ans[u] = min(a)
            for v in gr[u]:
                ans[v]=sr[v]-sr[par[u]]-ans[u]
                par[v] = u
                ne.append(v)
        else:
            for v in gr[u]:
                ans[v]=0
                par[v]=u
                ne.append(v)
    q = ne

b = 0

for i in range(len(ans)):
    if ans[i]<0:
        b = 1
#print(ans)
if b==1:
    print(-1)
else:
    print(sum(ans))
