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;
long long int XpowerY(long long int x, long long int y, long long int m) {
  long long int ans = 1;
  x = x % m;
  while (y > 0) {
    if (y % 2 == 1) ans = (ans * x) % m;
    x = ((x % m) * (x % m)) % m;
    y = y >> 1;
  }
  return ans % m;
}
vector<long long int> gr[100023];
long long int s[100023], val[100023];
long long int fl, minP;
long long int dfs(long long int node, long long int prev, long long int lev) {
  long long int mx = 1e12;
  if (lev % 2 == 0) {
    for (int i = 0; i < gr[node].size(); i++) {
      if (s[gr[node][i]] < prev) fl = 1;
      mx = min(mx, s[gr[node][i]] - prev);
    };
    val[node] = mx;
    if (gr[node].size() == 0) val[node] = 0;
  }
  for (int i = 0; i < gr[node].size(); i++) {
    val[gr[node][i]] = s[gr[node][i]] - mx - prev;
    long long int temp;
    if (lev % 2 == 0)
      temp = prev + val[node];
    else
      temp = s[node];
    dfs(gr[node][i], temp, lev + 1);
  }
  return 0;
}
int main() {
  long long int n, x;
  cin >> n;
  for (long long int i = 0; i < n - 1; i++) {
    cin >> x;
    gr[x].push_back(i + 2);
  }
  for (long long int i = 1; i <= n; i++) cin >> s[i];
  dfs(1, -1, 1);
  val[1] = s[1];
  if (fl) return cout << -1 << endl, 0;
  for (long long int i = 0; i < n; i++) {
    ;
    minP += val[i + 1];
  }
  cout << minP << endl;
  return 0;
}
