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 in[100007], res = 0, flag = 1;
vector<long long> adj[100007];
void dfs(long long s, long long pre_sum) {
  if (flag == 0) return;
  long long val = (in[s] == -1) ? pre_sum : in[s];
  if (pre_sum > val) {
    flag = 0;
    return;
  }
  if (in[s] == -1) {
    long long mn = 1e9;
    for (long long i = 0; i < adj[s].size(); i++) {
      long long val1 = adj[s][i];
      val1 = in[val1];
      val1 = (val1 == -1) ? 0 : val1;
      mn = min(mn, val1);
    }
    if (mn == 1e9) mn = 0;
    val = max(val, mn);
  }
  long long cal = val - pre_sum;
  res += cal;
  for (long long i = 0; i < adj[s].size(); i++) {
    dfs(adj[s][i], val);
  }
}
int main() {
  long long n, a, i;
  cin >> n;
  for (i = 0; i < n - 1; i++) {
    cin >> a;
    adj[a].push_back(i + 2);
  }
  for (i = 0; i < n; i++) cin >> in[i + 1];
  dfs(1, 0);
  res = (flag == 0) ? -1 : res;
  cout << res << "\n";
  return 0;
}
