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 TAM = 2e5 + 5;
int n;
int parent[TAM], acum[TAM], value[TAM], anc[TAM];
vector<int> sons[TAM];
vector<int> adj[TAM];
long long ans = 0;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  int x;
  parent[1] = 1;
  for (int i = 2; i <= n; i++) {
    cin >> parent[i];
    sons[parent[i]].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    cin >> acum[i];
    if (acum[i] == -1) {
      for (int v : sons[i]) {
        adj[parent[i]].push_back(v);
        anc[v] = parent[i];
      }
    }
  }
  value[1] = acum[1];
  queue<int> bfs;
  bfs.push(1);
  ans = acum[1];
  while (bfs.size()) {
    int curr = bfs.front();
    bfs.pop();
    if (curr != 1) {
      if (acum[anc[curr]] > acum[curr]) {
        cout << -1 << endl;
        return 0;
      }
      value[curr] = acum[curr] - acum[anc[curr]];
      ans += value[curr];
    }
    for (int v : adj[curr]) {
      bfs.push(v);
    }
  }
  for (int i = n; i > 0; i--) {
    if (acum[i] != -1 or sons[i].size() == 0) continue;
    int mn = INT_MAX;
    for (int w : sons[i]) mn = min(mn, value[w]);
    ans -= (mn * (sons[i].size() - 1));
  }
  cout << ans << endl;
  return 0;
}
