Appleman has a tree with n vertices. Some of the vertices (at least one) are colored black and other vertices are colored white.

Consider a set consisting of k (0 ≤ k < n) edges of Appleman's tree. If Appleman deletes these edges from the tree, then it will split into (k + 1) parts. Note, that each part will be a tree with colored vertices.

Now Appleman wonders, what is the number of sets splitting the tree in such a way that each resulting part will have exactly one black vertex? Find this number modulo 1000000007 (109 + 7).

Input

The first line contains an integer n (2 ≤ n ≤ 105) — the number of tree vertices. 

The second line contains the description of the tree: n - 1 integers p0, p1, ..., pn - 2 (0 ≤ pi ≤ i). Where pi means that there is an edge connecting vertex (i + 1) of the tree and vertex pi. Consider tree vertices are numbered from 0 to n - 1.

The third line contains the description of the colors of the vertices: n integers x0, x1, ..., xn - 1 (xi is either 0 or 1). If xi is equal to 1, vertex i is colored black. Otherwise, vertex i is colored white.

Output

Output a single integer — the number of ways to split the tree modulo 1000000007 (109 + 7).

Examples

Input

3
0 0
0 1 1


Output

2


Input

6
0 1 1 0 4
1 1 0 0 1 0


Output

1


Input

10
0 1 2 1 4 4 4 0 8
0 0 0 1 0 1 1 0 0 1


Output

27
import java.io.*;
import java.math.*;
import java.util.*;


public class SolutionD {
    public static void main(String[] args){
        new SolutionD().run();
    }
    ArrayList<Integer> arr[];
    int n;
    int mod = 1000000007;
    int col[];
    long dp[][];
    void dfs(int x){
        dp[x][0] = col[x];
        dp[x][1] = 1-col[x];
        for(int c: arr[x]) {
            dfs(c);
            dp[x][0] = (dp[x][1] *  dp[c][0] + dp[x][0] * dp[c][0] + dp[x][0] * dp[c][1]) % mod;
            dp[x][1] = ((dp[c][0] + dp[c][1]) * 1L * dp[x][1]) % mod; 
        }
    }
    void solve(){
        n = in.nextInt();
        arr = new ArrayList[n];
        dp = new long[n][2];
        for(int i = 0; i < n; i++) arr[i]=new ArrayList<Integer>();
        for(int i = 0, x; i < n-1; i++){
            x = in.nextInt();
            arr[x].add(i+1);
        }
        col = in.nextArray(n);
        dfs(0);
        out.println(dp[0][0]);
    }
    PrintWriter out;
    FastScanner in;

    public void run(){

        in = new FastScanner(System.in);
        out = new PrintWriter(System.out);
        solve();
        out.close();
    }

    class FastScanner{
        BufferedReader bf;
        StringTokenizer st;
        public FastScanner(InputStream is){
            bf = new BufferedReader(new InputStreamReader(is));
        }
        String next(){
            while(st == null || !st.hasMoreTokens()){
                try{
                    st = new StringTokenizer(bf.readLine());
                }
                catch(Exception ex){
                    ex.printStackTrace();
                }
            }
            return st.nextToken();
        }
        int nextInt(){
            return Integer.parseInt(next());
        }
        long nextLong(){
            return Long.parseLong(next());
        }
        double nextDouble(){
            return Double.parseDouble(next());
        }
        float nextFloat(){
            return Float.parseFloat(next());
        }
        BigInteger nextBigInteger(){
            return new BigInteger(next());
        }
        BigDecimal nextBigDecimal(){
            return new BigDecimal(next());
        }   
        int[] nextArray(int n){
            int x[] = new int[n];
            for(int i = 0; i < n; i++){
                x[i] = Integer.parseInt(next());
            }
            return x;
        }
    }
}
