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>
const int MAX = 2 * 1e5 + 2, MOD = 1e9 + 7;
using namespace std;
int n, a[MAX], s[MAX], p, mn = 1e9;
vector<int> v[MAX];
void dfs(int root, int pa, int depth) {
  if (depth == 1)
    a[root] = s[root] - s[pa];
  else if (depth == 0 && v[root].size() == 1)
    a[root] = 0, s[root] = s[pa];
  else {
    mn = 1e9;
    for (int x : v[root]) {
      if (x != pa) mn = min(mn, s[x] - s[pa]);
    }
    a[root] = mn;
    s[root] = a[root] + s[pa];
  }
  if (v[root].size() > 0)
    for (int x : v[root]) {
      if (x != pa) dfs(x, root, 1 - depth);
    }
}
int main() {
  s[0] = 0;
  cin >> n;
  for (int i = 1; i <= n - 1; i++) {
    scanf("%i", &p);
    v[i + 1].push_back(p);
    v[p].push_back(i + 1);
  }
  for (int i = 1; i <= n; i++) scanf("%i", &s[i]);
  dfs(1, 0, 1);
  for (int i = 1; i <= n; i++)
    if (a[i] < 0) {
      cout << -1;
      return 0;
    }
  long long sum = 0;
  for (int i = 1; i <= n; i++) sum += (long long)a[i];
  cout << sum;
  return 0;
}
