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>
typedef struct vertex {
  int parent;
  int sum;
} vertex;
vertex v[100005];
int main() {
  int n;
  scanf("%d", &n);
  v[0].parent = -1;
  for (int i = 1; i < n; i++) {
    scanf("%d", &v[i].parent);
    v[i].parent--;
  }
  for (int i = 0; i < n; i++) {
    scanf("%d", &v[i].sum);
    if (i > 0 && v[i].sum != -1) {
      if (v[v[i].parent].sum == -1) {
        v[v[i].parent].sum = v[i].sum;
      } else if (v[v[i].parent].sum > v[i].sum) {
        v[v[i].parent].sum = v[i].sum;
      }
    }
  }
  for (int i = 1; i < n; i++) {
    if (v[i].sum == -1) {
      v[i].sum = v[v[i].parent].sum;
    } else if (v[i].sum < v[v[i].parent].sum) {
      printf("-1");
      return 0;
    }
  }
  long long result = v[0].sum;
  for (int i = 1; i < n; i++) {
    result += v[i].sum - v[v[i].parent].sum;
  }
  printf("%lld", result);
  return 0;
}
