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;
using adj_list = vector<vector<int>>;
using tree_d = vector<int>;
constexpr long long INF = LONG_LONG_MAX;
constexpr int ROOT = -1;
int main() {
  int n;
  cin >> n;
  tree_d p(n);
  p[0] = ROOT;
  adj_list g(n);
  for (int i = 1; i < n; ++i) {
    cin >> p[i];
    --p[i];
    g[p[i]].push_back(i);
  }
  vector<long long> s(n);
  for (int i = 0; i < n; ++i) {
    cin >> s[i];
  }
  vector<long long> a(n);
  a[0] = s[0];
  for (int u = 0; u < n; ++u) {
    if (s[u] != -1) continue;
    a[u] = INF;
    for (int v : g[u]) a[u] = min(a[u], s[v] - s[p[u]]);
    if (g[u].empty()) a[u] = 0;
    s[u] = s[p[u]] + a[u];
  }
  for (int u = 1; u < n; ++u) {
    a[u] = s[u] - s[p[u]];
  }
  if (*min_element(a.begin(), a.end()) < 0) {
    cout << -1;
  } else {
    long long a_total = accumulate(a.begin(), a.end(), 0LL);
    cout << a_total;
  }
  cout << '\n';
  return 0;
}
