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.io.*;
import java.util.*;
public class Code534B {

	static void dfs(int i , Vector<Vector<Integer>> v , boolean vis[], long s[] ,long a[])
	{
		if(!vis[i]) 
		{
			vis[i]=true ;
			Vector<Integer> ve = v.elementAt(i) ;
			for(int j : ve)
			{
				if(s[j]==-1) s[j]=s[i] ;
				a[j]=s[j]-s[i] ;
				dfs(j,v,vis,s,a) ;
			}
		}
	}
	public static void main(String[] args) throws IOException 
	{
		BufferedReader jk = new BufferedReader(new InputStreamReader(System.in)) ;

		StringTokenizer ana = new StringTokenizer(jk.readLine()) ;
		int n = Integer.parseInt(ana.nextToken()) ;
		Vector<Vector<Integer>> v = new Vector<Vector<Integer>>() ;
		long s[] = new long[n] ;
		long a[] = new long[n] ;
		ana = new StringTokenizer(jk.readLine()) ;
		for(int i=0 ; i<n ;i++)
		{
			v.add(new Vector<Integer>()) ;
		}
		for(int i=1 ;i<n ;i++)
		{
			int data=Integer.parseInt(ana.nextToken())-1 ;
			v.elementAt(data).add(i) ;
		}
		ana = new StringTokenizer(jk.readLine()) ;
		for(int i=0 ;i<n ;i++)
		{
			s[i]=Long.parseLong(ana.nextToken()) ;
		}
		boolean tem=false ; 

		if(s[0]==-1) s[0]=0 ;
		for(int i=0 ; i<n ;i++)
		{
			if(s[i]==-1)
			{
				Vector<Integer> ve= v.elementAt(i) ;
				if(ve.size()>0)
				{
					Vector<Long> f = new Vector<Long>() ;
					for(int j:ve) f.add(s[j]) ;
					long h = Collections.min(f) ;
					if(h!=-1) s[i]=h ; 
				}	
			}

		}
		a[0]=s[0] ;
		boolean vis[] = new boolean[n] ;
		dfs(0,v,vis,s,a) ;
		for(int i=0 ; i<n ;i++)
		{
			Vector<Integer> ve= v.elementAt(i) ;
			for(int j:ve)
			{
				if(s[j]<s[i])
				{
					tem=true ; break ;
				}
			}
		}
		if(tem)
		{
			System.out.println(-1);
		}
		else	
		{
			long res=0 ;
			for(int i=0 ; i<n ;i++) 
			{
				res+=a[i] ;
			}
			System.out.println(res);
		}





	}

}
