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;
struct tree {
  int par, a, sum, dep;
} tr[100010];
int f_dep(int i) {
  if (i == 1)
    return tr[i].dep = 1;
  else if (tr[i].dep)
    return tr[i].dep;
  else
    return tr[i].dep = 1 + f_dep(tr[i].par);
}
int main() {
  int n;
  long long x = 0;
  cin >> n;
  tr[1].par = -1;
  for (int i = 2; i <= n; i++) cin >> tr[i].par;
  for (int i = 1; i <= n; i++) {
    cin >> tr[i].sum, tr[i].a = 0x7fffffff, f_dep(i);
    if (tr[i].sum == -1) tr[i].sum = 0x7fffffff;
  }
  tr[1].a = tr[1].sum;
  for (int i = 2; i <= n; i++) {
    if (tr[i].dep & 1 == 1) {
      tr[tr[i].par].a =
          min(tr[tr[i].par].a, tr[i].sum - tr[tr[tr[i].par].par].sum),
      tr[tr[i].par].sum = min(tr[i].sum, tr[tr[i].par].sum);
    }
  }
  for (int i = 2; i <= n; i++) {
    if (tr[i].sum == 0x7fffffff) {
      tr[i].a = 0;
    } else
      tr[i].a = tr[i].sum - tr[tr[i].par].sum;
    if (tr[i].a < 0) {
      cout << -1;
      return 0;
    }
    x += tr[i].a;
  }
  x += tr[1].a;
  cout << x << endl;
}
