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;
long long ans = 0LL;
long long s[100077];
long long a[100077];
vector<int> b[100077];
bool ok = true;
void work(int l, int r) {
  if (a[l] < 0) {
    if (b[l].size() == 0)
      a[l] = 0;
    else {
      long long mini = s[b[l][0]];
      for (int i = 0; i < b[l].size(); i++) {
        mini = min(mini, s[b[l][i]]);
      }
      long long dif = mini - s[r];
      s[l] = s[r] + dif;
      a[l] = dif;
      for (int i = 0; i < b[l].size(); i++) {
        int to = b[l][i];
        a[to] = s[to] - s[l];
      }
    }
  }
  for (int i = 0; i < b[l].size(); i++) {
    work(b[l][i], l);
  }
}
int main() {
  cin >> n;
  for (int i = 1; i < n; i++) {
    int p;
    cin >> p;
    p--;
    b[p].push_back(i);
  }
  for (int i = 0; i < n; i++) {
    cin >> s[i];
    a[i] = -1;
  }
  a[0] = s[0];
  work(0, -1);
  for (int i = 0; i < n; i++) {
    ok &= a[i] >= 0LL;
    ans += a[i];
  }
  if (!ok) {
    cout << -1 << endl;
    return 0;
  }
  cout << ans << endl;
  return 0;
}
