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;
inline long long gcd(long long a, long long b) {
  a = ((a) < 0 ? -(a) : (a));
  b = ((b) < 0 ? -(b) : (b));
  while (b) {
    a = a % b;
    swap(a, b);
  }
  return a;
}
const long long inf = 2147383647;
const long long mod = 1000000007;
const double pi = 2 * acos(0.0);
const double eps = 1e-9;
int flag;
long long ans, ar[100005];
vector<int> v[100005];
void dfs(int idx, int par, long long carry) {
  int x;
  if (ar[idx] == -1) {
    long long val = mod;
    for (int i = 0; i < ((long long)(v[idx]).size()); i++) {
      x = v[idx][i];
      if (x == par) continue;
      val = min(val, ar[x]);
    }
    if (val < mod) {
      ans += val - carry;
      for (int i = 0; i < ((long long)(v[idx]).size()); i++) {
        x = v[idx][i];
        if (x == par) continue;
        dfs(x, idx, val);
      }
    }
  } else {
    ans += ar[idx] - carry;
    for (int i = 0; i < ((long long)(v[idx]).size()); i++) {
      x = v[idx][i];
      if (x == par) continue;
      dfs(x, idx, ar[idx]);
    }
  }
}
void f(int idx, int par, long long carry) {
  int x;
  if (ar[idx] == -1) {
    for (int i = 0; i < ((long long)(v[idx]).size()); i++) {
      x = v[idx][i];
      if (x == par) continue;
      f(x, idx, carry);
    }
  } else {
    if (carry > ar[idx]) {
      flag = 1;
      return;
    }
    for (int i = 0; i < ((long long)(v[idx]).size()); i++) {
      x = v[idx][i];
      if (x == par) continue;
      f(x, idx, ar[idx]);
    }
  }
}
int main() {
  int n, x;
  cin >> n;
  for (int i = 1; i <= n - 1; i++) {
    scanf("%d", &x);
    v[i + 1].push_back(x);
    v[x].push_back(i + 1);
  }
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &ar[i]);
  }
  f(1, -1, 0);
  if (flag) {
    printf("-1\n");
    return 0;
  }
  ans = ar[1];
  for (int i = 0; i < ((long long)(v[1]).size()); i++) {
    dfs(v[1][i], 1, ar[1]);
  }
  printf("%lld\n", ans);
  return 0;
}
