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() {
  int n, k;
  cin >> n;
  vector<vector<int>> t(n + 1);
  vector<long long> s(n + 1);
  for (int i = 2; i <= n; ++i) {
    cin >> k;
    t[k].push_back(i);
  }
  for (int j = 1; j <= n; ++j) {
    cin >> k;
    s[j] = k;
  }
  queue<int> q;
  q.push(1);
  long long sum = s[1];
  while (!q.empty()) {
    int f = q.front();
    q.pop();
    long long c = LONG_LONG_MAX;
    for (int x : t[f]) {
      if (s[x] == -1)
        s[x] = s[f];
      else if (s[x] - s[f] < 0) {
        cout << "-1" << endl;
        return 0;
      } else
        sum += s[x] - s[f];
      if (s[x] < c) c = s[x];
      q.push(x);
    }
    if (!t[f].empty()) sum -= (c - s[f]) * (t[f].size() - 1);
  }
  cout << sum << endl;
}
