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;
const int N = 1e5;
int aa[N + 2];
vector<int> adj[N + 2];
void dfs(int node, int h, int sm) {
  if (h % 2 == 0) {
    if (adj[node].size() == 0) {
      aa[node] = 0;
      return;
    }
    int mn = 2e9;
    for (auto x : adj[node]) mn = min(mn, aa[x]);
    if (mn < sm) cout << -1 << endl, exit(0);
    aa[node] = mn - sm;
  } else
    aa[node] -= sm;
  for (auto x : adj[node]) dfs(x, h + 1, sm + aa[node]);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 2; i <= n; i++) {
    int x;
    cin >> x;
    adj[x].push_back(i);
  }
  for (int i = 1; i <= n; i++) cin >> aa[i];
  dfs(1, 1, 0);
  long long sm = 0;
  for (int i = 1; i <= n; i++) sm += aa[i] * 1LL;
  cout << sm << endl;
  return 0;
}
