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
import java.util.*;
import java.io.*;
public class D {
   public static void main(String[] args) {
      Scanner sc = new Scanner(System.in);
      
      int N = sc.nextInt();
      
      int[] p = new int[N + 1];
      ArrayList<Integer>[] suc = new ArrayList[N + 1];
      p[1] = 1;
      
      for(int i = 0; i <= N; i++) {
         suc[i] = new ArrayList<Integer>();
      }
      
      for(int i = 2; i <= N; i++) {
         p[i] = sc.nextInt();
         suc[p[i]].add(i);
      }
      
      long[] s = new long[N + 1];
      for(int i = 1; i <= N; i++) {
         s[i] = sc.nextLong();
      }
      
      //it is impossible for the successors to have s-value less than their predecessors
      for(int i = 2; i <= N; i++) {
         if(s[i] != -1 && s[i] < s[p[i]]) {
            System.out.println(-1);
            return;
         }
      }
      
      long[] ans = new long[N + 1];
      ans[1] = s[1];
      for(int i = 2; i <= N; i++) {
         if(s[i] == -1) {
            if(suc[i].size() == 0) {
               ans[i] = 0;
            }
            else {
               long val = (long)1e9 + 5;
               for(int j = 0; j < suc[i].size(); j++) {
                  val = Math.min(val, s[suc[i].get(j)]);
               }
               s[i] = val;
               ans[i] = val - s[p[i]];
               if(ans[i] < 0) {
                  System.out.println(-1);
                  return;
               }
            }
         }
         else {
            ans[i] = s[i] - s[p[i]];
            if(ans[i] < 0) {
               System.out.println(-1);
               return;
            }
         }
      }
      
      //consider the case 1-2, 2-3, 2-4. (0, -1, 5, 5) u want node 2 to take 5.
      long total = 0;
      for(long x : ans)
         total += x;
      System.out.println(total);
   }
}