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;
int N;
vector<int> par;
vector<vector<int>> child;
vector<int64_t> S;
vector<int64_t> A;
void dfs(int i) {
  int64_t par_val = (par[i] != -1 ? S[par[i]] : 0);
  if (S[i] != -1) {
    A[i] = S[i] - par_val;
  } else {
    if (child[i].size() == 0)
      A[i] = 0;
    else {
      A[i] = LLONG_MAX;
      for (auto y : child[i]) A[i] = min(A[i], S[y] - par_val);
    }
    S[i] = par_val + A[i];
  }
  for (auto y : child[i]) dfs(y);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> N;
  par = vector<int>(N);
  par[0] = -1;
  child = vector<vector<int>>(N);
  for (int i = 1; i < N; i++) {
    cin >> par[i];
    par[i]--;
    child[par[i]].push_back(i);
  }
  S = vector<int64_t>(N);
  A = vector<int64_t>(N);
  for (int i = 0; i < N; i++) cin >> S[i];
  dfs(0);
  int64_t sum = 0;
  bool soln = true;
  for (int i = 0; i < N; i++) {
    sum += A[i];
    if (A[i] < 0) soln = false;
  }
  if (!soln) sum = -1;
  cout << sum << endl;
  return 0;
}
