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 m = 1e5;
vector<vector<int>> A(m);
int p[m];
int s[m];
void dfs(int x) {
  int minn = 1e9 + 1;
  for (int i = 0; i < A[x].size(); i++) {
    if (s[A[x][i]] < minn) minn = s[A[x][i]];
    dfs(A[x][i]);
  }
  if (s[x] == -1) {
    if (minn == 1e9 + 1)
      s[x] = s[p[x] - 1];
    else
      s[x] = minn;
  }
  return;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  p[0] = 1;
  for (int i = 1; i < n; i++) {
    cin >> p[i];
    A[p[i] - 1].push_back(i);
  }
  for (int i = 0; i < n; i++) cin >> s[i];
  dfs(0);
  long long summ = s[0];
  for (int i = 0; i < n; i++) {
    if (s[i] < s[p[i] - 1]) {
      cout << -1;
      return 0;
    } else
      summ += s[i] - s[p[i] - 1];
  }
  cout << summ;
}
