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;
vector<int> v[100010];
int s[100010];
long long ans = 0;
long long dfs(int x) {
  long long ret;
  if (v[x].size() == 0) {
    ret = s[x];
  } else {
    vector<long long> vec;
    long long mini = 0;
    for (int i = 0; i < v[x].size(); ++i) {
      vec.push_back(dfs(v[x][i]));
      if (vec.back() != -1) mini = vec.back();
    }
    if (s[x] != -1) {
      mini = s[x];
      for (int i = 0; i < vec.size(); ++i) {
        if (vec[i] == -1) {
          vec[i] = mini;
        } else if (vec[i] < mini) {
          printf("-1\n");
          exit(0);
        }
      }
    } else {
      for (int i = 0; i < vec.size(); ++i) {
        if (vec[i] == -1) continue;
        mini = min(mini, vec[i]);
      }
      for (int i = 0; i < vec.size(); ++i) {
        if (vec[i] == -1) {
          vec[i] = mini;
        }
      }
    }
    for (int i = 0; i < vec.size(); ++i) {
      ans += vec[i] - mini;
    }
    ret = mini;
  }
  return ret;
}
int main() {
  int n;
  cin >> n;
  for (int i = 2; i <= n; ++i) {
    int x;
    cin >> x;
    v[x].push_back(i);
  }
  for (int i = 1; i <= n; ++i) {
    int x;
    cin >> x;
    s[i] = x;
  }
  cout << dfs(1) + ans << endl;
}
