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
#include <bits/stdc++.h>
using namespace std;
int p[100009], A[100009];
vector<int> tree[100009];
long long ans = 0;
void go(int now) {
  if (A[now] != -1) {
    int dif = A[now] - A[p[now]];
    if (dif < 0) {
      puts("-1");
      exit(0);
    }
    ans += dif;
  } else {
    if (tree[now].size() == 0) return;
    int par = A[p[now]], mn = 1e9;
    for (auto& it : tree[now]) mn = min(mn, A[it] - par);
    if (mn < 0) {
      puts("-1");
      exit(0);
    }
    A[now] = par + mn;
    ans += mn;
  }
  for (auto& it : tree[now]) go(it);
}
int main() {
  int N;
  scanf("%d", &N);
  p[1] = 0;
  for (int i = 2; i <= N; i++) {
    scanf("%d", &p[i]);
    tree[p[i]].push_back(i);
  }
  for (int i = 1; i <= N; i++) scanf("%d", &A[i]);
  go(1);
  printf("%lld", ans);
  return 0;
}
