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;
vector<int> kids[100001];
void process(long long* s, long long* a, int vertex, int par) {
  if (s[vertex] == -1 && kids[vertex].size() == 0) s[vertex] = s[par];
  if (s[vertex] == -1 && kids[vertex].size()) {
    int minn = INT_MAX;
    for (int i = kids[vertex].size() - 1; i >= 0; i--) {
      int kid = kids[vertex][i];
      if (s[kid] < minn) minn = s[kid];
    }
    s[vertex] = minn;
  }
  for (int i = kids[vertex].size() - 1; i >= 0; i--) {
    process(s, a, kids[vertex][i], vertex);
  }
  a[vertex] = s[vertex] - s[par];
}
int main() {
  int n;
  cin >> n;
  long long* s = new long long[n + 1];
  s[0] = 0;
  for (int i = 2; i < n + 1; i++) {
    int par;
    cin >> par;
    kids[par].push_back(i);
  }
  for (int i = 1; i < n + 1; i++) cin >> s[i];
  long long* a = new long long[n + 1];
  process(s, a, 1, 0);
  bool isValid = true;
  long long sum = 0;
  for (int i = 1; i < n + 1; i++) {
    if (a[i] < 0) {
      isValid = false;
      break;
    } else
      sum += a[i];
  }
  if (isValid)
    cout << sum;
  else
    cout << -1;
  delete[] s;
  delete[] a;
  return 0;
}
