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;
vector<vector<int> > G;
vector<int> sums;
vector<int> av;
bool impossible;
void dfs(int v, int s) {
  if (sums[v] != -1) {
    if (s > sums[v]) {
      impossible = true;
      return;
    }
    av[v] = sums[v] - s;
  } else {
    int64_t mn = (1ll << 60);
    for (int k : G[v]) {
      if (s > sums[k]) {
        impossible = true;
        return;
      }
      if (sums[k] - s < mn) {
        mn = sums[k] - s;
      }
    }
    if (mn == (1ll << 60)) {
      av[v] = 0;
    } else
      av[v] = mn;
  }
  for (int k : G[v]) {
    dfs(k, av[v] + s);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> pred(n + 3, 0);
  sums.assign(n + 3, 0);
  av.assign(n + 3, 0);
  G.assign(n + 3, vector<int>());
  impossible = false;
  for (int i = (2), _b = (n); i <= _b; ++i) {
    cin >> pred[i];
  }
  for (int i = (1), _b = (n); i <= _b; ++i) {
    cin >> sums[i];
  }
  for (int i = (2), _b = (n); i <= _b; ++i) {
    G[pred[i]].push_back(i);
  }
  dfs(1, 0);
  if (impossible) {
    cout << -1 << endl;
  } else {
    int64_t sum = 0;
    for (int i = (1), _b = (n); i <= _b; ++i) {
      sum += av[i];
    }
    cout << sum << endl;
  }
}
