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;
const int maxn = 1e5 + 1;
vector<int> Tree[maxn];
vector<int> parent(maxn);
vector<long long int> a(maxn);
vector<long long int> s(maxn);
void dfs(int p, bool odd) {
  if (!odd) {
    long long int mm = 1e9 + 1;
    for (int q : Tree[p]) {
      mm = min(mm, s[q] - s[parent[p]]);
    }
    a[p] = mm;
    if (Tree[p].empty()) a[p] = 0;
    s[p] = s[parent[p]] + a[p];
    for (int q : Tree[p]) {
      a[q] = s[q] - s[p];
      dfs(q, true);
    }
  } else {
    for (int q : Tree[p]) {
      dfs(q, false);
    }
  }
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i < n; i++) {
    int p;
    cin >> p;
    parent[i] = p - 1;
    Tree[p - 1].push_back(i);
  }
  for (int i = 0; i < n; i++) {
    cin >> s[i];
  }
  a[0] = s[0];
  dfs(0, true);
  long long int sum = 0;
  for (int i = 0; i < n; i++) {
    if (a[i] < 0) {
      cout << -1 << endl;
      return 0;
    }
    sum += a[i];
  }
  cout << sum << endl;
  return 0;
}
