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 i = 0, j, k, p, m, n;
  cin >> n;
  bool sum_given[n];
  vector<int> parent(n), arr(n, 0), sum(n);
  vector<vector<int>> child(n);
  parent[0] = -1;
  for (i = 1; i < n; i++) {
    cin >> p;
    parent[i] = p - 1;
    child[p - 1].push_back(i);
  }
  for (i = 0; i < n; i++) {
    cin >> sum[i];
    sum_given[i] = sum[i] == -1 ? 0 : 1;
  }
  arr[0] = sum[0];
  for (i = 1; i < n; i++) {
    if (sum_given[i] == 0) {
      if (child[i].size() == 0) {
        arr[i] = 0;
        sum[i] = sum[parent[i]];
      } else {
        j = sum[child[i][0]];
        for (k = 1; k < child[i].size(); k++) {
          j = min(sum[child[i][k]], j);
        }
        p = sum[parent[i]];
        if (j < p) {
          cout << "-1";
          return 0;
        } else {
          sum[i] = j;
          arr[i] = j - p;
        }
      }
    }
  }
  for (i = 1; i < n; i++) {
    if (sum_given[i] == 1) {
      p = sum[parent[i]];
      if (sum[i] < p) {
        cout << "-1";
        return 0;
      } else {
        arr[i] = sum[i] - p;
      }
    }
  }
  long long a = 0;
  for (i = 0; i < n; i++) {
    a += arr[i];
  }
  cout << a;
  return 0;
}
