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<vector<long long int> > adj(100005, vector<long long int>());
vector<long long int> s;
vector<long long int> dp;
long long int dfs(long long int x) {
  long long int cnt = 1e18;
  for (long long int i = 0; i < adj[x].size(); i++) {
    cnt = min(cnt, dfs(adj[x][i]));
  }
  if (s[x] >= 0) {
    cnt = min(cnt, s[x]);
  }
  return dp[x] = cnt;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int n;
  cin >> n;
  for (long long int i = 2; i <= n; i++) {
    long long int par;
    cin >> par;
    adj[par].push_back(i);
  }
  dp.resize(n + 1, 0);
  s.resize(n + 1);
  for (long long int i = 1; i <= n; i++) {
    cin >> s[i];
  }
  dfs(1);
  bool check = true;
  long long int ans = 0;
  queue<pair<int, int> > q;
  q.push({1, 0});
  while (!q.empty()) {
    long long int f = q.front().first;
    long long int x = q.front().second;
    q.pop();
    if (s[f] == -1) {
      if (dp[f] != 1e18 and x < dp[f]) {
        ans = ans + (dp[f] - x);
        x = dp[f];
      }
    } else {
      if (x > s[f]) {
        check = false;
        break;
      } else {
        ans = ans + (s[f] - x);
        x = s[f];
      }
    }
    for (long long int i = 0; i < adj[f].size(); i++) {
      q.push({adj[f][i], x});
    }
  }
  if (check) {
    cout << ans << endl;
  } else {
    cout << -1 << endl;
  }
  return 0;
}
