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;
long long a[123456];
bool dfs(vector<vector<int>> &t, int u, vector<int> &s, int h, long long cur) {
  if (h % 2 == 1) {
    if (!t[u].empty()) a[u] = numeric_limits<int>::max();
    for (int v : t[u]) {
      if (s[v] == -1) {
        return false;
      }
      a[u] = min(a[u], s[v] - cur);
    }
  } else {
    if (s[u] == -1) {
      return false;
    }
    a[u] = s[u] - cur;
  }
  if (a[u] < 0) {
    return false;
  }
  for (int v : t[u]) {
    if (!dfs(t, v, s, h + 1, cur + a[u])) {
      return false;
    }
  }
  return true;
}
void solve(istream &in, ostream &out) {
  int n;
  in >> n;
  vector<vector<int>> t(n);
  for (int i = 1; i < n; ++i) {
    int p;
    in >> p;
    t[p - 1].push_back(i);
  }
  vector<int> s(n);
  for (int i = 0; i < n; ++i) {
    in >> s[i];
  }
  bool res = dfs(t, 0, s, 0, 0);
  if (!res) {
    out << -1 << endl;
  } else {
    out << accumulate(a, a + n, 0LL) << endl;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  std::istream &in(std::cin);
  std::ostream &out(std::cout);
  solve(in, out);
}
