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 main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, flag = 0;
  long long ans = 0;
  cin >> n;
  int p[n + 1], s[n + 1];
  vector<int> a[n + 1];
  for (int i = 2; i <= n; i++) {
    cin >> p[i];
    a[p[i]].push_back(i);
  }
  for (int i = 1; i <= n; i++) cin >> s[i];
  queue<pair<int, int>> q;
  q.push({1, 1});
  while (!q.empty()) {
    pair<int, int> p1 = q.front();
    q.pop();
    if (p1.second % 2 != 0) {
      if (p1.first == 1)
        ans += s[1];
      else
        ans += s[p1.first] - s[p[p1.first]];
    }
    int m = INT_MAX, flag1 = 0;
    for (int i = 0; i < a[p1.first].size(); i++) {
      q.push({a[p1.first][i], p1.second + 1});
      m = min(m, s[a[p1.first][i]]);
      flag1 = 1;
    }
    if (p1.second % 2 == 0 && flag1 == 1) {
      s[p1.first] = m;
      if (m < s[p[p1.first]]) {
        flag = 1;
        break;
      }
      ans += s[p1.first] - s[p[p1.first]];
    }
  }
  if (flag == 1)
    cout << "-1\n";
  else
    cout << ans << "\n";
  return 0;
}
