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 javax.print.DocFlavor;
import java.util.*;
import java.io.*;

public class Main {
    static ArrayList<Integer>graph[]=new ArrayList[120000];
    public static void main(String[] args) throws IOException{
        BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(new PrintStream(System.out));
        int n=Integer.parseInt(f.readLine());
        for(int i=0;i<n;i++){
            graph[i]=new ArrayList<>();
        }
        StringTokenizer st=new StringTokenizer(f.readLine());
        for(int i=0;i<n-1;i++){
            int a=Integer.parseInt(st.nextToken());
            graph[i+1].add(a);
            graph[a].add(i+1);
        }
        st=new StringTokenizer(f.readLine());
        for(int i=0;i<n;i++){
            arr[i]=Integer.parseInt(st.nextToken());
        }
        dfs(0,-1);
        System.out.println(dp[0][1]);
        f.close();
        out.close();
    }
    static long MOD=1000000007;
    static long[][]dp=new long[120000][2];
    static void dfs(int curr, int last){
        dp[curr][0]=1-arr[curr];
        dp[curr][1]=arr[curr];
        for(int i:graph[curr]){
            if(i!=last) {
                dfs(i, curr);

                long temp1 = dp[curr][0];
                long temp2 = dp[curr][1];
                dp[curr][0] = 0;
                dp[curr][1] = 0;
                //next is part of it
                dp[curr][1] += temp1 * dp[i][1];
                dp[curr][1] += temp2 * dp[i][0];
                //
                dp[curr][0] += temp1 * dp[i][1];
                dp[curr][1] += temp2 * dp[i][1];

                dp[curr][0] += temp1 * dp[i][0];
            }
            dp[curr][0]%=MOD;
            dp[curr][1]%=MOD;
        }

    }
    static int[]arr=new int[120000];
}
class pair implements Comparable <pair>{
    int num;
    int idx;

    public int compareTo(pair other){
        return num- other.num;
    }


    pair(int a, int b)
    {
        num=a;
        idx=b;
    }
}
