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 p[100002];
int n;
long long s[100002];
long long a[100002];
long long ans;
vector<int> children[100002];
bool possible = true;
void computeEven(int node, int depth) {
  if (depth) {
    for (int i = 0; i < children[node].size(); i++)
      computeEven(children[node][i], 0);
    return;
  }
  if (!children[node].size()) {
    a[node] = 0;
    return;
  }
  a[node] = 10000000000;
  for (int i = 0; i < children[node].size(); i++) {
    long long u = s[children[node][i]] - s[p[node]];
    a[node] = min(a[node], u);
    computeEven(children[node][i], 1);
  }
  if (a[node] < 0) {
    possible = false;
    return;
  }
  ans += a[node];
  s[node] = a[node] + s[p[node]];
}
void computeOdd(int node, int depth) {
  if (depth) {
    a[node] = s[node] - s[p[node]];
    if (a[node] < 0) {
      possible = false;
      return;
    }
    ans += a[node];
  }
  for (int i = 0; i < children[node].size(); i++)
    computeOdd(children[node][i], (depth + 1) % 2);
}
void imprime() {
  for (int i = 1; i < n; i++) cout << s[i] << " ";
  cout << s[n] << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 2; i <= n; i++) {
    cin >> p[i];
    children[p[i]].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    cin >> s[i];
  }
  computeEven(1, 1);
  if (possible) {
    computeOdd(1, 1);
    if (possible)
      cout << ans << endl;
    else
      cout << -1 << endl;
  } else
    cout << -1 << endl;
  return 0;
}
