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;
bool minCost(int* par, int* s, int* a, int n);
int main() {
  int n;
  cin >> n;
  int *par = new int[n + 1], *s = new int[n + 1], *a = new int[n + 1];
  for (int i = 0; i <= n; ++i) par[i] = s[i] = a[i] = 0;
  for (int i = 2; i <= n; ++i) cin >> par[i];
  for (int i = 1; i <= n; ++i) cin >> s[i];
  if (minCost(par, s, a, n)) {
    long long res = 0;
    for (int i = 1; i <= n; ++i) res += a[i];
    cout << res << endl;
  } else
    cout << -1 << endl;
  delete[] par;
  delete[] s;
  delete[] a;
  return 0;
}
bool minCost(int* par, int* s, int* a, int n) {
  a[1] = s[1];
  for (int i = 2; i <= n; ++i) {
    if (s[i] == -1) {
      s[i] = s[par[i]];
      int min = INT_MAX;
      for (int j = i + 1; j <= n; ++j)
        if (par[j] == i && s[j] < min) {
          if (s[j] < s[i]) return false;
          if (s[j] < min) min = s[j];
        }
      if (min != INT_MAX) s[i] = min;
      a[i] = s[i] - s[par[i]];
    } else {
      if (s[par[i]] > s[i]) return false;
      a[i] = s[i] - s[par[i]];
    }
  }
  return true;
}
