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;
using namespace std::chrono;
vector<int> s;
vector<vector<int> > tree;
int max_num = 1e9 + 1;
int fill_s(int node) {
  int minval = max_num;
  for (int i = 0; i < tree[node].size(); i++) {
    int child = tree[node][i];
    int s_child = fill_s(child);
    if (s[child] == -1) {
      s[child] = s_child;
    }
    minval = min(s[child], minval);
  }
  if (s[node] == -1) {
    s[node] = minval;
  }
  return s[node];
}
vector<int> get_a(int n) {
  vector<int> a(n);
  a[0] = s[0];
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < tree[i].size(); j++) {
      if (s[tree[i][j]] == max_num)
        a[tree[i][j]] = 0;
      else {
        a[tree[i][j]] = s[tree[i][j]] - s[i];
      }
    }
  }
  return a;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  auto start = high_resolution_clock::now();
  int n;
  cin >> n;
  tree.resize(n);
  s.resize(n);
  for (int i = 1; i < n; i++) {
    int temp;
    cin >> temp;
    tree[temp - 1].push_back(i);
  }
  for (int i = 0; i < n; i++) {
    int temp;
    cin >> temp;
    s[i] = temp;
  }
  int root = fill_s(0);
  if (root == max_num) {
    cout << 0;
  } else {
    vector<int> ans = get_a(n);
    long long int sum = 0;
    int flag = 0;
    for (int i = 0; i < n; i++) {
      sum += (long long int)(ans[i]);
      if (ans[i] < 0) {
        flag = 1;
        break;
      }
    }
    if (flag == 0)
      cout << sum;
    else
      cout << -1;
  }
  auto stop = high_resolution_clock::now();
  auto duration = duration_cast<microseconds>(stop - start);
  return 0;
}
