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() {
  long long int n;
  cin >> n;
  long long int parent[n + 1];
  long long int h[n + 1];
  vector<long long int> v[n + 1];
  for (long long int i = 2; i <= n; i++) {
    long long int temp;
    cin >> temp;
    parent[i] = temp;
    v[temp].push_back(i);
  }
  long long int sv[n + 1];
  long long int av[n + 1];
  for (long long int i = 1; i <= n; i++) {
    cin >> sv[i];
  }
  if (sv[1] == -1) {
    av[1] = 0;
    sv[1] = 0;
  } else
    av[1] = sv[1];
  queue<long long int> q;
  q.push(1);
  h[1] = 1;
  while (!q.empty()) {
    long long int curr = q.front();
    q.pop();
    if (curr != 1) {
      h[curr] = h[parent[curr]] + 1;
      long long int mini = INT_MAX;
      for (long long int i = 0; i < v[curr].size(); i++) {
        mini = min(sv[v[curr][i]], mini);
        q.push(v[curr][i]);
      }
      if (sv[curr] == -1) {
        if (mini - sv[parent[curr]] < 0) {
          cout << -1 << endl;
          exit(0);
        }
        if (v[curr].size() == 0 || mini == INT_MAX) {
          av[curr] = 0;
          sv[curr] = sv[parent[curr]];
        } else {
          av[curr] = mini - sv[parent[curr]];
          sv[curr] = sv[parent[curr]] + av[curr];
        }
      } else {
        av[curr] = sv[curr] - sv[parent[curr]];
      }
    } else {
      for (long long int i = 0; i < v[curr].size(); i++) {
        q.push(v[curr][i]);
      }
    }
  }
  long long int sum = 0;
  for (long long int i = 1; i <= n; i++) {
    sum += av[i];
  }
  cout << sum << endl;
}
