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;
const int maxN = 1e5 + 5;
vector<int> adj[maxN];
int s[maxN], par[maxN];
void dfs(int n, int p) {
  par[n] = p;
  for (int i = int(0); i < int((int)adj[n].size()); i++) {
    if (adj[n][i] != p) dfs(adj[n][i], n);
  }
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cin.tie(0), cout.tie(0),
      cout.precision(15);
  int n, tmp;
  cin >> n;
  for (int i = int(0); i < int(n - 1); i++) {
    cin >> tmp;
    tmp -= 1;
    adj[tmp].push_back(i + 1);
  }
  dfs(0, -1);
  for (int i = int(0); i < int(n); i++) {
    cin >> s[i];
  }
  long long ans = s[0];
  for (int i = int(0); i < int(n); i++) {
    if (s[i] == -1) {
      int n = i, tt = s[par[i]];
      int mini = 2000000000;
      for (int j = int(0); j < int((int)adj[n].size()); j++) {
        if (s[adj[n][j]] < tt) {
          cout << -1;
          return 0;
        }
        mini = min(mini, s[adj[n][j]] - tt);
      }
      if (mini != 2000000000) ans += mini;
      for (int j = int(0); j < int((int)adj[n].size()); j++) {
        ans += s[adj[n][j]] - tt - mini;
      }
    }
  }
  cout << ans;
}
