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;
bool f(int u, int p, vector<vector<int>> &adj, vector<long long> &s,
       long long &sum) {
  long long au = -1;
  if (s[u] == -1) {
    long long mins = LONG_MAX;
    for (int i = 0; i < adj[u].size(); i++) {
      int v = adj[u][i];
      if (v == p) {
        continue;
      }
      mins = min(mins, s[v]);
    }
    au = (mins == LONG_MAX ? 0 : mins - s[p]);
    s[u] = s[p] + au;
  } else {
    au = max(au, s[u] - (p == -1 ? 0 : s[p]));
  }
  if (au < 0) {
    return false;
  }
  for (int i = 0; i < adj[u].size(); i++) {
    int v = adj[u][i];
    if (v == p) {
      continue;
    }
    if (!f(v, u, adj, s, sum)) {
      return false;
    }
  }
  sum += au;
  return true;
}
int main() {
  int n;
  cin >> n;
  vector<vector<int>> adj(n + 1);
  for (int i = 2; i <= n; i++) {
    int p;
    cin >> p;
    adj[i].push_back(p);
    adj[p].push_back(i);
  }
  vector<long long> s(n + 1);
  for (int i = 1; i <= n; i++) {
    cin >> s[i];
  }
  long long sum = 0;
  bool found = f(1, -1, adj, s, sum);
  cout << (found ? sum : -1);
  return 0;
}
