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 main{

	static FastReader in = new FastReader();
	static PrintWriter out = new PrintWriter(System.out);
	static int N = (int)1e5+5;
	static ArrayList<Integer>[] ch = new ArrayList[N];
	static int[] p = new int[N];
	
	static void solve(){
		int n = in.nextInt();
		int[] av = new int[n+1];
		for(int i=1;i<=n;i++){
			ch[i] = new ArrayList<>();
			av[i] = Integer.MAX_VALUE;
		}
		for(int i=2;i<=n;i++){
			int x = in.nextInt();
			ch[x].add(i);
			p[i] = x;
		}
		int[] sv = new int[n+1];
		for(int i=1;i<=n;i++)sv[i] = in.nextInt();
		long ans = (long)sv[1], f = 1;
		
		Queue<Integer> q = new LinkedList<>();


		q.add(1);
		av[1] = sv[1];

		while(!q.isEmpty()){
			int x = q.poll();
			for(int k : ch[x])q.add(k);
			if(x==1)continue;
			if(sv[x]==-1){
				for(int k : ch[x])av[x] = Math.min(av[x],sv[k]-sv[p[x]]);
				if(ch[x].size()==0)av[x] = 0;
				if(av[x]<0){
					f = 0;
					break;
				}
				sv[x] = av[x]+sv[p[x]];
			}
			else{
				av[x] = sv[x]-sv[p[x]];
				f = sv[x]>=sv[p[x]]?f:0;
			}
			ans+=av[x];	
		}



		if(f==1)
			out.println(ans);
		else {
			out.println(-1);
		}
	}


	public static void main(String[] args){
		solve();
		out.close();
	}	

	static class FastReader{
        BufferedReader br;
        StringTokenizer st;
        public FastReader(){
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        public FastReader(String s) throws Exception{
            br = new BufferedReader(new FileReader(s));
        }

        String next(){
            while (st == null || !st.hasMoreElements()){
                try{
                    st = new StringTokenizer(br.readLine());
                }catch (IOException  e){
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        String nextLine(){
            String str = "";
            try{   
                str = br.readLine();
            }catch (IOException e){
                e.printStackTrace();
            }  
            return str;
        }
        int nextInt(){return Integer.parseInt(in.next());}
    	long nextLong(){return Long.parseLong(in.next());}
    	double nextDouble(){return Double.parseDouble(in.next());}
    }
}
