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 i, j, k, n, m;
  bool q = true;
  cin >> n;
  long long s = 0;
  int p[n];
  int arr[n];
  int h[n];
  h[0] = 1;
  p[0] = -1;
  for (i = 1; i < n; i++) {
    cin >> k;
    k--;
    p[i] = k;
    h[i] = h[k] + 1;
  }
  for (i = 0; i < n; i++) {
    cin >> k;
    if (p[i] < 0) {
      arr[i] = k;
    }
    if (h[i] % 2 == 1 && p[i] >= 0) {
      if (arr[p[i]] < 0) {
        arr[p[i]] = k;
        arr[i] = k;
        continue;
      }
      arr[p[i]] = min(arr[p[i]], k);
      arr[i] = k;
    }
    if (h[i] % 2 == 0) {
      arr[i] = k;
    }
  }
  for (i = 1; i < n; i++) {
    if (h[i] % 2 == 1 && p[i] >= 0) {
      s += arr[i] - arr[p[i]];
    }
    if (h[i] % 2 == 0) {
      if (arr[i] < 0) {
        continue;
      }
      if (arr[i] < arr[p[i]]) {
        q = false;
      } else {
        s += arr[i] - arr[p[i]];
      }
    }
  }
  s += arr[0];
  if (!q) {
    cout << "-1" << endl;
    return 0;
  }
  cout << s << endl;
}
