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;
vector<long long> v[100005];
long long arr[100005];
map<long long, long long> marked1, marked2;
void dfs1(long long idx, long long level, long long par_val) {
  marked1[idx] = 1;
  long long count = 0;
  long long min_val = INT_MAX;
  for (auto x : v[idx]) {
    if (marked1[x] == 1) {
      continue;
    }
    min_val = min(min_val, arr[x]);
    count++;
  }
  if (level % 2 == 1) {
    if (count == 0) {
      arr[idx] = par_val;
    } else {
      arr[idx] = min_val;
    }
  }
  for (auto x : v[idx]) {
    if (marked1[x] == 1) {
      continue;
    }
    dfs1(x, level + 1, arr[idx]);
  }
}
long long dfs2(long long idx, long long par_val) {
  marked2[idx] = 1;
  if (arr[idx] < par_val) {
    return -1;
  }
  long long sum = 0;
  for (auto x : v[idx]) {
    if (marked2[x] == 1) {
      continue;
    }
    long long y = dfs2(x, arr[idx]);
    if (y == -1) {
      return y;
    }
    sum += y;
  }
  return sum + arr[idx] - par_val;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  t = 1;
  while (t--) {
    long long n;
    cin >> n;
    long long i;
    for (i = 2; i < n + 1; i++) {
      long long a;
      cin >> a;
      v[a].push_back(i);
      v[i].push_back(a);
    }
    for (i = 1; i < n + 1; i++) {
      cin >> arr[i];
    }
    dfs1(1, 0, 0);
    {}
    long long ans = dfs2(1, 0);
    cout << ans << endl;
  }
  return 0;
}
