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 main() {
  int n;
  cin >> n;
  int a[n + 1], b[n + 1];
  for (int i = 2; i < n + 1; ++i) scanf("%d", &a[i]);
  scanf("%d", &b[1]);
  for (int i = 2; i < n + 1; ++i) {
    scanf("%d", &b[i]);
    if (b[i] != -1)
      if (b[a[i]] == -1 || b[a[i]] > b[i]) b[a[i]] = b[i];
  }
  for (int i = 2; i < n + 1; ++i) {
    if (b[i] == -1) b[i] = b[a[i]];
  }
  long long int sum = (long long int)b[1];
  for (int i = 2; i < n + 1; ++i) {
    if (b[i] < b[a[i]]) {
      printf("%d\n", -1);
      exit(0);
    } else
      sum += b[i] - b[a[i]];
  }
  printf("%lld\n", sum);
}
