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;
long long int pr[500100], su[500100];
long long int ans;
vector<int> v[500100];
void soll(int node, int p, int nd, int cnt) {
  cnt %= 2;
  if (cnt == 1 && node != 1) {
    if (su[node] - nd < 0) {
      cout << -1;
      exit(0);
    }
  }
  if (cnt == 0 && v[node].size() == 1) {
    su[node] = su[p];
    return;
  }
  long long int minn = 1e18;
  for (auto nd : v[node]) {
    if (nd == p) continue;
    minn = min(minn, su[nd]);
  }
  if (cnt == 0) su[node] = minn;
  for (auto nd : v[node]) {
    if (nd == p) continue;
    soll(nd, node, su[p], cnt + 1);
  }
}
void sol(int node, int p) {
  for (auto nd : v[node]) {
    if (nd == p) continue;
    ans += su[nd] - su[node];
    if (su[nd] - su[node] < 0) {
      cout << -1;
      exit(0);
    }
    sol(nd, node);
  }
}
int main() {
  int n;
  cin >> n;
  for (int i = 2; i <= n; i++) {
    int x;
    cin >> x;
    v[x].push_back(i);
    v[i].push_back(x);
  }
  for (int i = 1; i <= n; i++) cin >> su[i];
  soll(1, 0, 0, 1);
  sol(1, 0);
  ans += su[1];
  cout << ans;
  return 0;
}
