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;
template <typename T>
T Abs(T first) {
  return (first < 0 ? -first : first);
}
template <typename T>
T Sqr(T first) {
  return (first * first);
}
string plural(string s) {
  return (int((s).size()) && s[int((s).size()) - 1] == 'x' ? s + "en"
                                                           : s + "s");
}
const int INF = (int)1e9;
const long double EPS = 1e-12;
const long double PI = acos(-1.0);
bool Read(int &first) {
  char c, r = 0, n = 0;
  first = 0;
  for (;;) {
    c = getchar();
    if ((c < 0) && (!r)) return (0);
    if ((c == '-') && (!r))
      n = 1;
    else if ((c >= '0') && (c <= '9'))
      first = first * 10 + c - '0', r = 1;
    else if (r)
      break;
  }
  if (n) first = -first;
  return (1);
}
bool ReadLL(long long &first) {
  char c, r = 0, n = 0;
  first = 0;
  for (;;) {
    c = getchar();
    if ((c < 0) && (!r)) return (0);
    if ((c == '-') && (!r))
      n = 1;
    else if ((c >= '0') && (c <= '9'))
      first = first * 10 + c - '0', r = 1;
    else if (r)
      break;
  }
  if (n) first = -first;
  return (1);
}
vector<int> ch[200005];
int V[200005];
long long rec(int i, long long s) {
  int j, c;
  long long ret = 0;
  int v = 0;
  if (V[i] >= 0)
    v = V[i];
  else if (int((ch[i]).size())) {
    v = INF;
    for (j = 0; j < int((ch[i]).size()); j++) v = min(v, V[ch[i][j]]);
  } else
    v = s;
  if (s > v)
    return (-1);
  else {
    ret += v - s;
    s = v;
  }
  for (j = 0; j < int((ch[i]).size()); j++) {
    c = ch[i][j];
    long long r = rec(c, s);
    if (r < 0) return (-1);
    ret += r;
  }
  return (ret);
}
int main() {
  if (0) freopen("in.txt", "r", stdin);
  int N, M, K;
  int i, j, k, a, b, c, first, second, z;
  Read(N);
  for (i = 1; i <= N - 1; i++) Read(j), ch[j - 1].push_back(i);
  for (i = 0; i < N; i++) Read(V[i]);
  cout << rec(0, 0) << endl;
  return (0);
}
