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;
int n;
int p[300000];
int s[300000], t[300000];
vector<int> c[300000];
void solve() {
  cin >> n;
  for (int i = 1; i < n; ++i) {
    cin >> p[i];
    p[i]--;
    c[p[i]].push_back(i);
  }
  for (int i = 0; i < n; ++i) {
    cin >> s[i];
    t[i] = 1000000005;
  }
  t[0] = s[0];
  for (int i = 1; i < n; ++i) {
    if (s[i] == -1) continue;
    t[i] = s[i];
    t[p[i]] = min(t[p[i]], s[i]);
  }
  vector<int> active(c[0].begin(), c[0].end());
  long long res = s[0];
  while (!active.empty()) {
    int i = active.back();
    active.pop_back();
    if (s[i] == -1 && c[i].empty()) continue;
    for (int j : c[i]) active.push_back(j);
    if (t[i] < t[p[i]]) {
      cout << -1 << endl;
      return;
    }
    res += t[i] - t[p[i]];
  }
  cout << res << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cout << setprecision(12) << fixed;
  solve();
  return 0;
}
