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;
double EPS = 1e-9;
long long INF = 1000000000000000005ll;
double PI = acos(-1);
inline string IntToString(long long int a) {
  char x[100];
  sprintf(x, "%lld", a);
  string s = x;
  return s;
}
inline long long int StringToInt(string a) {
  char x[100];
  long long int res;
  strcpy(x, a.c_str());
  sscanf(x, "%lld", &res);
  return res;
}
inline string uppercase(string s) {
  int n = (int)s.size();
  for (long long int i = 0; i < n; i++)
    if (s[i] >= 'a' && s[i] <= 'z') s[i] = s[i] - 'a' + 'A';
  return s;
}
inline string lowercase(string s) {
  int n = (int)s.size();
  for (long long int i = 0; i < n; i++)
    if (s[i] >= 'A' && s[i] <= 'Z') s[i] = s[i] - 'A' + 'a';
  return s;
}
bool prime(long long int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
bool cmp(pair<long long int, long long int> a,
         pair<long long int, long long int> b) {
  long long int as = a.first - a.second;
  long long int binary_search = b.first - b.second;
  return (as > binary_search);
}
bool vis[200005];
vector<long long int> v[200005];
long long int a[200005];
long long int s[200005];
bool b;
void dfs(long long int u, long long int par) {
  vis[u] = 1;
  for (int child : v[u]) {
    if (vis[child] == 0) {
      if (s[child] == -1) {
        long long int mi = INT_MAX;
        for (int cc : v[child]) {
          if (s[cc] != -1 && vis[cc] == 0) {
            mi = min(s[cc], mi);
          }
        }
        if (mi < s[u]) {
          s[child] = s[u];
          a[child] = 0;
        } else if (mi != INT_MAX) {
          s[child] = mi;
          a[child] = s[child] - s[u];
        } else {
          s[child] = s[u];
          a[child] = 0;
        }
      } else if (s[child] < s[u]) {
        b = 1;
      } else {
        a[child] = s[child] - s[u];
      }
      dfs(child, u);
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int n;
  cin >> n;
  for (long long int i = 2; i <= n; i++) {
    long long int x;
    cin >> x;
    v[i].push_back(x);
    v[x].push_back(i);
  }
  for (long long int i = 1; i <= n; i++) {
    long long int x;
    cin >> x;
    s[i] = x;
  }
  a[1] = s[1];
  dfs(1, 0);
  if (b) {
    cout << -1;
    return 0;
  }
  long long int sum = 0;
  for (long long int i = 1; i <= n; i++) {
    sum += a[i];
  }
  cout << sum << '\n';
  return 0;
}
