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 parent[100001];
int flag = 0;
void dfs(vector<long long int> v[], int n, long long int s[], long long int a[],
         int source, long long int h[], bool visited[]) {
  visited[source] = true;
  if (h[source] % 2 != 0) {
    for (int i = 0; i < v[source].size(); i++) {
      h[v[source][i]] = h[source] + 1;
      dfs(v, n, s, a, v[source][i], h, visited);
    }
  } else {
    long long int m = INT_MAX;
    long long y = s[source];
    if (v[source].size() == 0) {
      a[source] = 0;
    } else {
      for (int i = 0; i < v[source].size(); i++) {
        h[v[source][i]] = h[source] + 1;
        long long int q = v[source][i];
        m = min(m, s[q]);
        if (s[q] < s[parent[source]]) {
          flag = 1;
          break;
        }
      }
      a[source] = m - s[parent[source]];
      for (int i = 0; i < v[source].size(); i++) {
        long long int q = v[source][i];
        a[q] = s[q] - m;
        dfs(v, n, s, a, v[source][i], h, visited);
      }
    }
  }
}
int main() {
  int n;
  cin >> n;
  vector<long long int> v[n + 1];
  for (int i = 2; i <= n; i++) {
    long long p;
    cin >> p;
    parent[i] = p;
    v[p].push_back(i);
  }
  long long s[n + 1], a[n + 1];
  for (int i = 1; i <= n; i++) cin >> s[i];
  a[1] = s[1];
  bool visited[n + 1];
  long long int h[n + 1];
  h[1] = 1;
  memset(visited, n + 1, false);
  dfs(v, n, s, a, 1, h, visited);
  if (flag == 1)
    cout << "-1" << endl;
  else {
    long long sum = 0;
    for (int i = 1; i <= n; i++) sum = sum + a[i];
    cout << sum << endl;
  }
}
