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() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int n;
  cin >> n;
  vector<long long int> p(n, -1);
  map<long long int, vector<long long int> > c;
  c[0] = vector<long long int>();
  vector<long long int> s(n, 0);
  for (long long int x = 1; x < n; x++) {
    c[x] = vector<long long int>();
    long long int n;
    cin >> n;
    n--;
    p[x] = n;
    c[n].push_back(x);
  }
  for (long long int x = 0; x < n; x++) {
    cin >> s[x];
  }
  for (long long int x = n - 1; x >= 0; x--) {
    if (s[x] == -1) {
      long long int mn = INT_MAX;
      for (long long int i = 0; i < ((int)(c[x]).size()); i++) {
        mn = min(mn, s[c[x][i]]);
      }
      if (mn == INT_MAX) {
        mn = s[p[x]];
      }
      s[x] = mn;
    }
  }
  long long int sum = s[0];
  for (long long int i = 1; i < n; i++) {
    if (s[i] < s[p[i]]) {
      cout << -1 << endl;
      return 0;
    }
    sum = sum + s[i] - s[p[i]];
  }
  cout << sum << endl;
  return 0;
}
