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 gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
bool mod(long double a, long double b) { return a / b - floor(a / b); }
int occurs(vector<int> v, int n) {
  auto it = lower_bound(((v)).begin(), ((v)).end(), (n));
  auto it1 = upper_bound(((v)).begin(), ((v)).end(), (n));
  int x = it1 - it;
  return x;
}
long double max(long double a, long double b) { return a < b ? b : a; }
const int N = 1e5 + 1;
vector<int> g[N], p, s, a;
long long ans = 0;
bool f = 0;
void dfs(int v) {
  for (int u : g[v]) {
    if (u == p[v]) continue;
    if (s[u] == -1) {
      int mn = 1e9 + 1, k = 0;
      long long sum = 0;
      for (int i : g[u]) {
        if (i == p[u]) continue;
        if (s[i] == -1) continue;
        sum += s[i];
        k++;
        mn = min(mn, s[i]);
      }
      if (k) s[u] = min(mn, (int)(sum / k));
      s[u] = max(s[u], s[v]);
      a[u] = s[u] - s[v];
    } else {
      if (s[u] < s[v])
        f = 1;
      else {
        a[u] = s[u] - s[v];
      }
    }
    dfs(u);
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  a = s = p = vector<int>(n);
  for (int(i) = 1; (i) < (n); ++(i)) {
    int x;
    cin >> x;
    x--;
    p[i] = x;
    g[i].push_back(x);
    g[x].push_back(i);
  }
  for (int(i) = 0; (i) < (n); ++(i)) {
    cin >> s[i];
  }
  a[0] = s[0];
  dfs(0);
  if (f) {
    cout << -1;
    return 0;
  }
  for (int(i) = 0; (i) < (n); ++(i)) {
    ans += a[i];
  }
  cout << ans;
  return 0;
}
