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
import sys

n = int(input())
sys.setrecursionlimit(100*1000+100)

ch = [[] for _ in range(n)]
for i, x in enumerate(map(int, input().split(' ')), start=1):
    ch[x-1].append(i)

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

ans = 0
def dfs(v, buf):
    global ans
    while True:
        m = min((s[c] for c in ch[v]), default=buf) if s[v] == -1 else s[v]
        if m < buf: ans = -1
        if ans == -1: return
        ans += m - buf
        if len(ch[v]) == 1:
            v = ch[v][0]
            buf = m
            continue
        else:
            for c in ch[v]: 
                dfs(c, m)
            break

dfs(0, 0)
print (ans)