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 *
n=int(input())
fa=[0,0]+[int(i) for i in input().split()]
s=[0]+[int(i) for i in input().split()]
g=[[] for i in range(n+1)]
for x in range(2,len(fa)):
    g[fa[x]].append(x)

sub=[0]*(n+1)
q=deque([1])
rcr=[]
while(q):
    x=q.popleft()
    rcr.append(x)
    for y in g[x]:
        q.append(y)
rcr.reverse()

inf=int(1e10)
s1=[inf if i==-1 else i for i in s]
for x in rcr:
    sub[x]=s1[x]
    if g[x]:
        sub[x]=min(sub[x],  min([sub[y] for y in g[x]]) )

dp=[0]*(n+1)
a=[0]*(n+1)
q=deque(g[1])
a[1] = s[1] if s[1]!=-1 else 0
dp[1]=a[1]
ans=a[1]
ok=1
while(q):
    x=q.popleft()
    if sub[x]==inf:
        a[x]=0
    else:
        if s[x]==-1 : a[x]=sub[x]-dp[fa[x]]
        else : a[x]=s[x]-dp[fa[x]]
    if a[x]<0 :
        ok=0
        break
    dp[x]=dp[fa[x]]+a[x]
    ans+=a[x]
    for y in g[x]:
        q.append(y)

print(ans) if ok else print(-1)     
